300 PROBLEMS IN SPECIAL AND

GENERAL RELATIVITY
Einstein's theories of special relativity and general relativity form a core part of today's undergraduate (or master's-level) physics curriculum. This is a supplementary problem book or student's manual, consisting of 150 problems in each of special and general relativity i.e., in total 300 problems. The problems have been collected, developed, tested, and refined by the authors over the past two decades from homework and exams given at KTH Royal Institute of Technology, Stockholm, Sweden, starting in the late 1990s. They are a mixture of short-form and multipart extended problems, with hints provided where appropriate. Complete solutions are elaborated for every problem, in a different section of the book; some solutions include brief discussions on their physical or historical significance. The extensive and fully worked out solutions are the main feature of the book and have been revised several times by the authors. Designed as a companion text to complement a main relativity textbook, it does not assume access to any specific textbook. This is a helpful resource for advanced students, for self-study, as a source of problems for university teaching assistants, or as an inspiration for instructors and examiners constructing problems for their lectures, homework, or exams.
mattias blennow is Associate Professor in Theoretical Astroparticle Physics at KTH Royal Institute of Technology, Stockholm, Sweden. His research is mainly directed toward the physics of neutrinos and dark matter and physics beyond the Standard Model. He is the author of the textbook Mathematical Methods for Physics and Engineering (CRC Press, 2018). He has more than 15 years of experience in teaching and has taught special and general relativity both as a lecturer and as a teaching assistant.
tommy ohlsson is Professor of Theoretical Physics at KTH Royal Institute of Technology, Stockholm, Sweden. He has also been a visiting professor at the University of Iceland, Reykjavik, Iceland, over several years. His main research field is theoretical particle physics, especially neutrino physics and physics beyond the Standard Model. He has written the textbook Relativistic Quantum Physics: From Advanced Quantum Mechanics to Introductory Quantum Field Theory (Cambridge University Press, 2011). He has more than 25 years of university-level teaching experience in relativity theory and physics in general.

300 PROBLEMS IN SPECIAL AND GENERAL RELATIVITY

With Complete Solutions
MATTIAS BLENNOW
KTH Royal Institute of Technology
TOMMY OHLSSON
KTH Royal Institute of Technology

CAMERRDGE


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Names: Blennow, Mattias, 1980- author | Ohlsson, Tommy, 1973- author.
Title: 300 problems in special and general relativity : with complete solutions / M. Blennow & T. Ohlsson.
Other titles: Three hundred problems in special and general relativity Description: New York : Cambridge University Press, [2021]|
Includes bibliographical references and index.
Identifiers: LCCN 2021017342 (print) | LCCN 2021017343 (ebook) |
ISBN 9781316510674 (hardback) | ISBN 9781009017732 (paperback) | ISBN 9781009039345 (epub)
Subjects: LCSH: Special relativity (Physics)-Problems, exercises, etc. |
General relativity (Physics)-Problems, exercises, etc. | LCGFT: Problems and exercises.
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To our wives, Ana and Linda

Contents

Preface page ix
Notation, Concepts, and Conventions in Relativity Theory ..... 1
General Notation ..... 1
Special Relativity ..... 3
General Relativity ..... 6
Conventions ..... 12
1 Special Relativity Theory ..... 14
1.1 Basics ..... 14
1.2 Length Contraction, Time Dilation, and Spacetime Diagrams ..... 15
1.3 Lorentz Transformations and Geometry of Minkowski Space ..... 20
1.4 Relativistic Velocities and Proper Quantities ..... 22
1.5 Relativistic Optics ..... 26
1.6 Relativistic Mechanics ..... 29
1.7 Electromagnetism ..... 36
1.8 Energy-Momentum Tensor ..... 41
1.9 Lagrange's Formalism ..... 42
2 General Relativity Theory ..... 43
2.1 Some Differential Geometry ..... 43
2.2 Christoffel Symbols, Riemann and Ricci Tensors, and Einstein's Equations ..... 49
2.3 Maxwell's Equations and Energy-Momentum Tensor ..... 53
2.4 Killing Vector Fields ..... 55
2.5 Schwarzschild Metric ..... 57
2.6 Metrics, Geodesic Equations, and Proper Quantities ..... 58
2.7 Kruskal-Szekeres Coordinates ..... 66
2.8 Weak Field Approximation and Newtonian Limit ..... 68
2.9 Gravitational Lensing ..... 70
2.10 Frequency Shifts ..... 70
2.11 Gravitational Waves ..... 72
2.12 Cosmology and Friedmann-Lemaître-Robertson-Walker Metric ..... 74
3 Solutions to Problems ..... 77
3.1 Solutions to Problems in Special Relativity Theory ..... 77
3.2 Solutions to Problems in General Relativity Theory ..... 185
Bibliography ..... 350
Index to the Problems and Solutions ..... 352

Preface

This book is a supplementary book in the form of a "problem book" or "student's manual" in special and general relativity consisting of a total of 300 problems ( 150 problems each in special and general relativity) with complete and elaborate solutions. It is intended as a companion text to a main textbook, but does not assume any particular textbook. It may be used for self-study act as a source of problems for classes, or as inspiration for teachers and examiners looking to construct new problems for lectures, homework, or exams.
The problems have been collected over the course of about two decades from homework and exams given at KTH Royal Institute of Technology, Stockholm, Sweden, starting in the late 1990s. The extensive and fully worked-out solutions are the main feature of the book and have been revised several times by the authors.
The book is divided into the following chapters:
"Notation, Concepts, and Conventions in Relativity Theory";
  1. Problems in "Special Relativity Theory";
  2. Problems in "General Relativity Theory"; and
  3. "Solutions to Problems" in both special and general relativity,
    where the first, unnumbered chapter introduces and sets the stage for both special and general relativity and is intended to be a brief review. The structure of the book is to first present the problems belonging to each main chapter (i.e., Chapters 1 and 2), which are further split into sections in order to obtain a better overview. The solutions are then presented in Chapter 3 (i.e., they do not follow immediately after the problem formulations). The main purpose of this is to suppress the urge for the reader to look at the solution to a problem before making a proper attempt. Some of the problems and solutions are illustrated by figures.
The target audience of the book is students and teachers of special and/or general relativity courses at the master's level that may benefit from it in the way described
above. It will generally be too advanced for the relativity covered by the typical introductory modern physics courses at the bachelor's level, and most likely not advanced enough for an in-depth study at the PhD level.
Finally, we would like to acknowledge our colleagues Jouko Mickelsson, Håkan Snellman, Edwin Langmann, and Teresia Månsson, who have given important contributions to some of the problem statements included in this book. We would also like to thank our editor, Vince Higgs, at Cambridge University Press for a smooth and constructive process with the publication of this book, Torbjörn Bäck for supporting us in developing this book, and Marcus Pernow for proofreading earlier versions of the problem statements and solutions in special relativity. In addition, the KTH Royal Institute of Technology in Stockholm, Sweden, and the University of Iceland in Reykjavik, Iceland, are acknowledged for their hospitality and financial support.

Notation, Concepts, and Conventions in Relativity Theory

This chapter serves to briefly review the concepts relevant to the problems presented in this book. Its purpose is to remind the reader of the basic concepts as well as to introduce the notations and conventions that will be used. In particular, some notations and conventions will vary throughout the different textbooks available on the subject. Some of the different notations have been deliberately used in a number of problems in order to familiarize the reader with the fact that different notations occur in the literature.

General Notation

The components of a vector V V VVV will be written as V μ V μ V^(mu)V^{\mu}Vμ in contravariant form and V μ V μ V_(mu)V_{\mu}Vμ in covariant form with the index μ μ mu\muμ running over all the spacetime coordinates. When referring to 3 -vectors, Latin letters will be used for the spatial indices rather than Greek ones, which we use for spacetime coordinates. In the case when an explicit basis for a given vector space is needed, we will use the partial derivatives μ μ del_(mu)\partial_{\mu}μ to denote such a basis, i.e.,
(0.1) V = V μ μ (0.1) V = V μ μ {:(0.1)V=V^(mu)del_(mu):}\begin{equation*} V=V^{\mu} \partial_{\mu} \tag{0.1} \end{equation*}(0.1)V=Vμμ
Similarly, tensor components will be denoted with superscripts for contravariant indices and subscripts for covariant indices. Tensors with n n nnn indices, all down, are called covariant tensors of rank n n nnn and tensors with n n nnn indices, all up, are called contravariant tensors of rank n n nnn. Tensors with indices both up and down are so-called mixed tensors. Thus, a vector is a tensor of rank 1 (one index) and a scalar is a tensor of rank zero (no indices). The Einstein summation convention is used throughout
the book, implying that indices that are repeated are to be summed over the relevant range. For example, in a four-dimensional spacetime, we have
(0.2) V μ U μ μ = 0 3 V μ U μ = V 0 U 0 + V 1 U 1 + V 2 U 2 + V 3 U 3 (0.2) V μ U μ μ = 0 3 V μ U μ = V 0 U 0 + V 1 U 1 + V 2 U 2 + V 3 U 3 {:(0.2)V^(mu)U_(mu)-=sum_(mu=0)^(3)V^(mu)U_(mu)=V^(0)U_(0)+V^(1)U_(1)+V^(2)U_(2)+V^(3)U_(3):}\begin{equation*} V^{\mu} U_{\mu} \equiv \sum_{\mu=0}^{3} V^{\mu} U_{\mu}=V^{0} U_{0}+V^{1} U_{1}+V^{2} U_{2}+V^{3} U_{3} \tag{0.2} \end{equation*}(0.2)VμUμμ=03VμUμ=V0U0+V1U1+V2U2+V3U3
where U U UUU and V V VVV are vectors. Contravariant and covariant components are related by lowering and raising with the metric tensor g = ( g μ ν ) g = g μ ν g=(g_(mu nu))g=\left(g_{\mu \nu}\right)g=(gμν) and its inverse g 1 = ( g μ ν ) g 1 = g μ ν g^(-1)=(g^(mu nu))g^{-1}=\left(g^{\mu \nu}\right)g1=(gμν), respectively, i.e.,
(0.3) V μ = g μ ν V ν and V μ = g μ ν V v (0.3) V μ = g μ ν V ν  and  V μ = g μ ν V v {:(0.3)V_(mu)=g_(mu nu)V^(nu)quad" and "quadV^(mu)=g^(mu nu)V_(v):}\begin{equation*} V_{\mu}=g_{\mu \nu} V^{\nu} \quad \text { and } \quad V^{\mu}=g^{\mu \nu} V_{v} \tag{0.3} \end{equation*}(0.3)Vμ=gμνVν and Vμ=gμνVv
Partial derivatives of a given function f f fff may be denoted in several ways, e.g.,
(0.4) f x μ = μ f = f , μ (0.4) f x μ = μ f = f , μ {:(0.4)(del f)/(delx^(mu))=del_(mu)f=f_(,mu):}\begin{equation*} \frac{\partial f}{\partial x^{\mu}}=\partial_{\mu} f=f_{, \mu} \tag{0.4} \end{equation*}(0.4)fxμ=μf=f,μ
Several indices after the comma in the latter notation represent higher-order derivatives and the notation may also be used for vector components, for which indices belonging to the vector component are written before the comma and indices denoting derivatives after the comma, i.e.,
(0.5) μ ν f = f , μ ν and μ V ν = V ν , μ (0.5) μ ν f = f , μ ν  and  μ V ν = V ν , μ {:(0.5)del_(mu)del_(nu)f=f_(,mu nu)quad" and "quaddel_(mu)V_(nu)=V_(nu,mu):}\begin{equation*} \partial_{\mu} \partial_{\nu} f=f_{, \mu \nu} \quad \text { and } \quad \partial_{\mu} V_{\nu}=V_{\nu, \mu} \tag{0.5} \end{equation*}(0.5)μνf=f,μν and μVν=Vν,μ
Objects with two indices may be represented in matrix form. We will indicate this by putting parentheses around the considered objects. For example, we can write the object A A AAA with two indices as
(0.6) A = ( A μ ν ) = ( A 00 A 01 A 02 A 03 A 10 A 11 A 12 A 13 A 20 A 21 A 22 A 23 A 30 A 31 A 32 A 33 ) (0.6) A = A μ ν = A 00 A 01 A 02 A 03 A 10 A 11 A 12 A 13 A 20 A 21 A 22 A 23 A 30 A 31 A 32 A 33 {:(0.6)A=(A_(mu nu))=([A_(00),A_(01),A_(02),A_(03)],[A_(10),A_(11),A_(12),A_(13)],[A_(20),A_(21),A_(22),A_(23)],[A_(30),A_(31),A_(32),A_(33)]):}A=\left(A_{\mu \nu}\right)=\left(\begin{array}{llll} A_{00} & A_{01} & A_{02} & A_{03} \tag{0.6}\\ A_{10} & A_{11} & A_{12} & A_{13} \\ A_{20} & A_{21} & A_{22} & A_{23} \\ A_{30} & A_{31} & A_{32} & A_{33} \end{array}\right)(0.6)A=(Aμν)=(A00A01A02A03A10A11A12A13A20A21A22A23A30A31A32A33)
By convention, the first index represents the row of the matrix and the second index represents the column. When this is used for one covariant index and one contravariant index, the contravariant index is taken as the row index and the covariant index as the column index.
For objects with more than two indices, we may use matrix notation to represent parts of such objects by inserting a bullet (' \bullet ') in place of the indices being considered. For example, the components A 1 ν μ A 1 ν μ A_(1nu)^(mu)A_{1 \nu}^{\mu}A1νμ of the three-index object A λ ν μ A λ ν μ A_(lambda nu)^(mu)A_{\lambda \nu}^{\mu}Aλνμ would be represented as the matrix
(0.7) A 1 = ( A 1 v μ ) = ( A 10 0 A 11 0 A 12 0 A 13 0 A 10 1 A 11 1 A 12 1 A 13 1 A 10 2 A 11 2 A 12 2 A 13 2 A 10 3 A 11 3 A 12 3 A 13 3 ) . (0.7) A 1 = A 1 v μ = A 10 0 A 11 0 A 12 0 A 13 0 A 10 1 A 11 1 A 12 1 A 13 1 A 10 2 A 11 2 A 12 2 A 13 2 A 10 3 A 11 3 A 12 3 A 13 3 . {:(0.7)A_(1∙)^(∙)=(A_(1v)^(mu))=([A_(10)^(0),A_(11)^(0),A_(12)^(0),A_(13)^(0)],[A_(10)^(1),A_(11)^(1),A_(12)^(1),A_(13)^(1)],[A_(10)^(2),A_(11)^(2),A_(12)^(2),A_(13)^(2)],[A_(10)^(3),A_(11)^(3),A_(12)^(3),A_(13)^(3)]).:}A_{1 \bullet}^{\bullet}=\left(A_{1 v}^{\mu}\right)=\left(\begin{array}{cccc} A_{10}^{0} & A_{11}^{0} & A_{12}^{0} & A_{13}^{0} \tag{0.7}\\ A_{10}^{1} & A_{11}^{1} & A_{12}^{1} & A_{13}^{1} \\ A_{10}^{2} & A_{11}^{2} & A_{12}^{2} & A_{13}^{2} \\ A_{10}^{3} & A_{11}^{3} & A_{12}^{3} & A_{13}^{3} \end{array}\right) .(0.7)A1=(A1vμ)=(A100A110A120A130A101A111A121A131A102A112A122A132A103A113A123A133).

Special Relativity

In a flat 1 + 3 1 + 3 1+31+31+3-dimensional spacetime and in Cartesian coordinates, the Minkowski metric is given by
(0.8) d s 2 = η μ ν d x μ d x ν = c 2 d t 2 d x 2 d y 2 d z 2 (0.8) d s 2 = η μ ν d x μ d x ν = c 2 d t 2 d x 2 d y 2 d z 2 {:(0.8)ds^(2)=eta_(mu nu)dx^(mu)dx^(nu)=c^(2)dt^(2)-dx^(2)-dy^(2)-dz^(2):}\begin{equation*} d s^{2}=\eta_{\mu \nu} d x^{\mu} d x^{\nu}=c^{2} d t^{2}-d x^{2}-d y^{2}-d z^{2} \tag{0.8} \end{equation*}(0.8)ds2=ημνdxμdxν=c2dt2dx2dy2dz2
where c c ccc is the speed of light in vacuum and x 0 = c t x 0 = c t x^(0)=ctx^{0}=c tx0=ct. In units of c = 1 c = 1 c=1c=1c=1, so-called natural units, it holds that x 0 = t x 0 = t x^(0)=tx^{0}=tx0=t. The metric tensor and its inverse, i.e., the inverse metric tensor, can be written as
η = ( η μ ν ) = ( 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ) η 1 = ( η μ ν ) = ( 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ) η = η μ ν = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 η 1 = η μ ν = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 eta=(eta_(mu nu))=([1,0,0,0],[0,-1,0,0],[0,0,-1,0],[0,0,0,-1])quad<=>quadeta^(-1)=(eta^(mu nu))=([1,0,0,0],[0,-1,0,0],[0,0,-1,0],[0,0,0,-1])\eta=\left(\eta_{\mu \nu}\right)=\left(\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{array}\right) \quad \Leftrightarrow \quad \eta^{-1}=\left(\eta^{\mu \nu}\right)=\left(\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{array}\right)η=(ημν)=(1000010000100001)η1=(ημν)=(1000010000100001).
For any two vectors x = ( x μ ) = ( x 0 , x 1 , x 2 , x 3 ) x = x μ = x 0 , x 1 , x 2 , x 3 x=(x^(mu))=(x^(0),x^(1),x^(2),x^(3))x=\left(x^{\mu}\right)=\left(x^{0}, x^{1}, x^{2}, x^{3}\right)x=(xμ)=(x0,x1,x2,x3) and y = ( y v ) = ( y 0 , y 1 , y 2 , y 3 ) y = y v = y 0 , y 1 , y 2 , y 3 y=(y^(v))=(y^(0),y^(1),y^(2),y^(3))y=\left(y^{v}\right)=\left(y^{0}, y^{1}, y^{2}, y^{3}\right)y=(yv)=(y0,y1,y2,y3) in Minkowski space described by their contravariant components expressed in Cartesian coordinates, the Minkowski inner product is introduced as
(0.10) x y x 0 y 0 x 1 y 1 x 2 y 2 x 3 y 3 (0.10) x y x 0 y 0 x 1 y 1 x 2 y 2 x 3 y 3 {:(0.10)x*y-=x^(0)y^(0)-x^(1)y^(1)-x^(2)y^(2)-x^(3)y^(3):}\begin{equation*} x \cdot y \equiv x^{0} y^{0}-x^{1} y^{1}-x^{2} y^{2}-x^{3} y^{3} \tag{0.10} \end{equation*}(0.10)xyx0y0x1y1x2y2x3y3
which is obviously commutative, i.e., x y = y x x y = y x x*y=y*xx \cdot y=y \cdot xxy=yx. We also define the notation x 2 x x = ( x 0 ) 2 ( x 1 ) 2 ( x 2 ) 2 ( x 3 ) 2 x 2 x x = x 0 2 x 1 2 x 2 2 x 3 2 x^(2)-=x*x=(x^(0))^(2)-(x^(1))^(2)-(x^(2))^(2)-(x^(3))^(2)x^{2} \equiv x \cdot x=\left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}x2xx=(x0)2(x1)2(x2)2(x3)2 for the squared norm ('length') of the vector x x xxx, which is indefinite, since it can be either positive or negative. 1 1 ^(1){ }^{1}1 The Minkowski metric η η eta\etaη and its inverse η 1 η 1 eta^(-1)\eta^{-1}η1 fulfill the relation
(0.11) η μ λ η λ v = η μ v = η μ ν = δ μ v , (0.11) η μ λ η λ v = η μ v = η μ ν = δ μ v , {:(0.11)eta_(mu lambda)eta^(lambda v)=eta_(mu)^(v)=eta_(mu)^(nu)=delta_(mu)^(v)",":}\begin{equation*} \eta_{\mu \lambda} \eta^{\lambda v}=\eta_{\mu}^{v}=\eta_{\mu}^{\nu}=\delta_{\mu}^{v}, \tag{0.11} \end{equation*}(0.11)ημληλv=ημv=ημν=δμv,
where δ μ v δ μ v delta_(mu)^(v)\delta_{\mu}^{v}δμv is the Kronecker delta such that δ μ v = 1 δ μ v = 1 delta_(mu)^(v)=1\delta_{\mu}^{v}=1δμv=1 if μ = v μ = v mu=v\mu=vμ=v and δ μ v = 0 δ μ v = 0 delta_(mu)^(v)=0\delta_{\mu}^{v}=0δμv=0 if μ v μ v mu!=v\mu \neq vμv. We can write the Minkowski inner product in multiple ways as
(0.12) x y = x μ η μ v y v = η μ ν x μ y v = x μ y μ = x 0 y 0 + x i y i = x 0 y 0 x i y i (0.12) x y = x μ η μ v y v = η μ ν x μ y v = x μ y μ = x 0 y 0 + x i y i = x 0 y 0 x i y i {:(0.12)x*y=x^(mu)eta_(mu v)y^(v)=eta_(mu nu)x^(mu)y^(v)=x^(mu)y_(mu)=x^(0)y_(0)+x^(i)y_(i)=x^(0)y^(0)-x^(i)y^(i):}\begin{equation*} x \cdot y=x^{\mu} \eta_{\mu v} y^{v}=\eta_{\mu \nu} x^{\mu} y^{v}=x^{\mu} y_{\mu}=x^{0} y_{0}+x^{i} y_{i}=x^{0} y^{0}-x^{i} y^{i} \tag{0.12} \end{equation*}(0.12)xy=xμημvyv=ημνxμyv=xμyμ=x0y0+xiyi=x0y0xiyi
where, e.g., x μ x μ x^(mu)x^{\mu}xμ can be considered as the contravariant components of the vector x x xxx and y μ y μ y_(mu)y_{\mu}yμ the covariant components of the vector y y yyy, i.e., y 0 = y 0 y 0 = y 0 y_(0)=y^(0)y_{0}=y^{0}y0=y0 and y i = y i y i = y i y_(i)=-y^(i)y_{i}=-y^{i}yi=yi, and it also holds that x μ y μ = x v y ν x μ y μ = x v y ν x^(mu)y_(mu)=x_(v)y^(nu)x^{\mu} y_{\mu}=x_{v} y^{\nu}xμyμ=xvyν. Furthermore, we say that the vector x x xxx is timelike if x 2 > 0 x 2 > 0 x^(2) > 0x^{2}>0x2>0, lightlike if x 2 = 0 x 2 = 0 x^(2)=0x^{2}=0x2=0, and spacelike if x 2 < 0 x 2 < 0 x^(2) < 0x^{2}<0x2<0. Note that lightlike vectors x x xxx form a cone ( x 0 ) 2 = ( x 1 ) 2 + ( x 2 ) 2 + ( x 3 ) 2 x 0 2 = x 1 2 + x 2 2 + x 3 2 (x^(0))^(2)=(x^(1))^(2)+(x^(2))^(2)+(x^(3))^(2)\left(x^{0}\right)^{2}=\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}+\left(x^{3}\right)^{2}(x0)2=(x1)2+(x2)2+(x3)2 and a nonspacelike vector x x xxx is future pointing if x 0 > 0 x 0 > 0 x^(0) > 0x^{0}>0x0>0 and past pointing if x 0 < 0 x 0 < 0 x^(0) < 0x^{0}<0x0<0.
In general, a Lorentz transformation Λ Λ Lambda\LambdaΛ between two coordinate systems S S SSS and S S S^(')S^{\prime}S described by coordinates x x xxx and x x x^(')x^{\prime}x, respectively, is given by
(0.13) x = Λ x x μ = Λ ν μ x ν (0.13) x = Λ x x μ = Λ ν μ x ν {:(0.13)x^(')=Lambda x quad<=>quadx^('mu)=Lambda_(nu)^(mu)x^(nu):}\begin{equation*} x^{\prime}=\Lambda x \quad \Leftrightarrow \quad x^{\prime \mu}=\Lambda_{\nu}^{\mu} x^{\nu} \tag{0.13} \end{equation*}(0.13)x=Λxxμ=Λνμxν
In particular, if the Lorentz transformation is a boost in the x 1 x 1 x^(1)x^{1}x1-direction, we can write
(0.14) Λ ( 01 ) = ( cosh θ sinh θ 0 0 sinh θ cosh θ 0 0 0 0 1 0 0 0 0 1 ) = ( γ β γ 0 0 β γ γ 0 0 0 0 1 0 0 0 0 1 ) (0.14) Λ ( 01 ) = cosh θ sinh θ 0 0 sinh θ cosh θ 0 0 0 0 1 0 0 0 0 1 = γ β γ 0 0 β γ γ 0 0 0 0 1 0 0 0 0 1 {:(0.14)Lambda^((01))=([cosh theta,-sinh theta,0,0],[-sinh theta,cosh theta,0,0],[0,0,1,0],[0,0,0,1])=([gamma,-beta gamma,0,0],[-beta gamma,gamma,0,0],[0,0,1,0],[0,0,0,1]):}\Lambda^{(01)}=\left(\begin{array}{cccc} \cosh \theta & -\sinh \theta & 0 & 0 \tag{0.14}\\ -\sinh \theta & \cosh \theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)=\left(\begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)(0.14)Λ(01)=(coshθsinhθ00sinhθcoshθ0000100001)=(γβγ00βγγ0000100001)
where θ θ theta\thetaθ is the rapidity, β v / c β v / c beta-=v//c\beta \equiv v / cβv/c, and γ γ gamma\gammaγ is the so-called gamma factor, i.e., γ γ ( v ) ( 1 v 2 / c 2 ) 1 / 2 γ γ ( v ) 1 v 2 / c 2 1 / 2 gamma-=gamma(v)-=(1-v^(2)//c^(2))^(-1//2)\gamma \equiv \gamma(v) \equiv\left(1-v^{2} / c^{2}\right)^{-1 / 2}γγ(v)(1v2/c2)1/2, with v v vvv being the relative speed between the coordinate systems S S SSS and S S S^(')S^{\prime}S. Furthermore, it holds that
cosh θ = γ = 1 1 v 2 / c 2 , sinh θ = β γ = v c 1 1 v 2 / c 2 , tanh θ = β = v c cosh θ = γ = 1 1 v 2 / c 2 , sinh θ = β γ = v c 1 1 v 2 / c 2 , tanh θ = β = v c cosh theta=gamma=(1)/(sqrt(1-v^(2)//c^(2))),quad sinh theta=beta gamma=(v)/(c)(1)/(sqrt(1-v^(2)//c^(2))),quad tanh theta=beta=(v)/(c)\cosh \theta=\gamma=\frac{1}{\sqrt{1-v^{2} / c^{2}}}, \quad \sinh \theta=\beta \gamma=\frac{v}{c} \frac{1}{\sqrt{1-v^{2} / c^{2}}}, \quad \tanh \theta=\beta=\frac{v}{c}coshθ=γ=11v2/c2,sinhθ=βγ=vc11v2/c2,tanhθ=β=vc.
The formulas for (Lorentz) length contraction and time dilation are given by
(0.16) = γ ( v ) = 1 v 2 / c 2 , t = t γ ( v ) = t 1 v 2 / c 2 (0.16) = γ ( v ) = 1 v 2 / c 2 , t = t γ ( v ) = t 1 v 2 / c 2 {:(0.16)ℓ^(')=(ℓ)/(gamma(v))=ℓsqrt(1-v^(2)//c^(2))","quad t=t^(')gamma(v)=(t^('))/(sqrt(1-v^(2)//c^(2))):}\begin{equation*} \ell^{\prime}=\frac{\ell}{\gamma(v)}=\ell \sqrt{1-v^{2} / c^{2}}, \quad t=t^{\prime} \gamma(v)=\frac{t^{\prime}}{\sqrt{1-v^{2} / c^{2}}} \tag{0.16} \end{equation*}(0.16)=γ(v)=1v2/c2,t=tγ(v)=t1v2/c2
respectively.
The relativistic energy-momentum dispersion relation is given by
(0.17) E = m γ ( v ) c 2 = m c 2 1 v 2 / c 2 (0.17) E = m γ ( v ) c 2 = m c 2 1 v 2 / c 2 {:(0.17)E^(')=m gamma(v)c^(2)=(mc^(2))/(sqrt(1-v^(2)//c^(2))):}\begin{equation*} E^{\prime}=m \gamma(v) c^{2}=\frac{m c^{2}}{\sqrt{1-v^{2} / c^{2}}} \tag{0.17} \end{equation*}(0.17)E=mγ(v)c2=mc21v2/c2
where m m mmm is the mass of an object, v v vvv its speed, and E E EEE its energy. In the rest frame of the object, it leads to Einstein's famous formula
(0.18) E = m c 2 (0.18) E = m c 2 {:(0.18)E=mc^(2):}\begin{equation*} E=m c^{2} \tag{0.18} \end{equation*}(0.18)E=mc2
The relativistic addition of velocities for colinear velocities v v vvv and v v v^(')v^{\prime}v is given by
(0.19) v = v + v 1 + v v / c 2 (0.19) v = v + v 1 + v v / c 2 {:(0.19)v^('')=(v+v^('))/(1+vv^(')//c^(2)):}\begin{equation*} v^{\prime \prime}=\frac{v+v^{\prime}}{1+v v^{\prime} / c^{2}} \tag{0.19} \end{equation*}(0.19)v=v+v1+vv/c2
In the nonrelativistic limit, i.e., v , v c v , v c v,v^(')≪cv, v^{\prime} \ll cv,vc, the classical formula v v + v v v + v v^('')≃v+v^(')v^{\prime \prime} \simeq v+v^{\prime}vv+v is recovered.
Consider radiation of light in a specific coordinate direction of the coordinate system S S SSS. One should think of the radiation as coming from a fixed source in this coordinate system, where the radiation has frequency v v vvv. For an observer in a
coordinate system S S S^(')S^{\prime}S moving along the same coordinate direction with the relative velocity v v vvv, a frequency v v v^(')v^{\prime}v is observed that is given by the relativistic Doppler formula, i.e.,
(0.20) v = v c v c + v (0.20) v = v c v c + v {:(0.20)v^(')=vsqrt((c-v)/(c+v)):}\begin{equation*} v^{\prime}=v \sqrt{\frac{c-v}{c+v}} \tag{0.20} \end{equation*}(0.20)v=vcvc+v
If the observer is moving away from the source, there is a redshift in the frequency of light, whereas if the observer is moving toward the source, there is a corresponding blueshift.
For the (binary) reaction A + B a + b + A + B a + b + A+B longrightarrow a+b+cdotsA+B \longrightarrow a+b+\cdotsA+Ba+b+, where two particles with 4-momenta p A p A p_(A)p_{A}pA and p B p B p_(B)p_{B}pB collide, using conservation of 4-momentum, we have
(0.21) p A + p B = p a + p b + , (0.21) p A + p B = p a + p b + , {:(0.21)p_(A)+p_(B)=p_(a)+p_(b)+cdots",":}\begin{equation*} p_{A}+p_{B}=p_{a}+p_{b}+\cdots, \tag{0.21} \end{equation*}(0.21)pA+pB=pa+pb+,
whereas for the decay A a + b + A a + b + A longrightarrow a+b+cdotsA \longrightarrow a+b+\cdotsAa+b+, we have the simpler relation
(0.22) p A = p a + p b + p c + (0.22) p A = p a + p b + p c + {:(0.22)p_(A)=p_(a)+p_(b)+p_(c)+cdots:}\begin{equation*} p_{A}=p_{a}+p_{b}+p_{c}+\cdots \tag{0.22} \end{equation*}(0.22)pA=pa+pb+pc+
This can be generalized to any number of particles with corresponding 4-momenta before and after a reaction, i.e.,
(0.23) P in = i = A , B , p i = j = a , b , p j = P out . (0.23) P in  = i = A , B , p i = j = a , b , p j = P out  . {:(0.23)P_("in ")=sum_(i=A,B,dots)p_(i)=sum_(j=a,b,dots)p_(j)=P_("out ").:}\begin{equation*} P_{\text {in }}=\sum_{i=A, B, \ldots} p_{i}=\sum_{j=a, b, \ldots} p_{j}=P_{\text {out }} . \tag{0.23} \end{equation*}(0.23)Pin =i=A,B,pi=j=a,b,pj=Pout .
Note that P 2 P 2 P^(2)P^{2}P2 is invariant for any P = k = 1 N p k P = k = 1 N p k P=sum_(k=1)^(N)p_(k)P=\sum_{k=1}^{N} p_{k}P=k=1Npk, where N N NNN is the number of particles, and actually, for any two 4 -vectors A A AAA and B B BBB, the Minkowski inner product A B = η μ ν A μ B ν = A μ B μ A B = η μ ν A μ B ν = A μ B μ A*B=eta_(mu nu)A^(mu)B^(nu)=A^(mu)B_(mu)A \cdot B=\eta_{\mu \nu} A^{\mu} B^{\nu}=A^{\mu} B_{\mu}AB=ημνAμBν=AμBμ is invariant under Lorentz transformations. Especially, A 2 = A μ A μ A 2 = A μ A μ A^(2)=A^(mu)A_(mu)A^{2}=A^{\mu} A_{\mu}A2=AμAμ is invariant. This is useful in many applications.
In electromagnetism, the electromagnetic field strength tensor F F FFF is defined as
(0.24) F μ ν = μ A v ν A μ , (0.24) F μ ν = μ A v ν A μ , {:(0.24)F^(mu nu)=del^(mu)A^(v)-del^(nu)A^(mu)",":}\begin{equation*} F^{\mu \nu}=\partial^{\mu} A^{v}-\partial^{\nu} A^{\mu}, \tag{0.24} \end{equation*}(0.24)Fμν=μAvνAμ,
where A = ( A μ ) = ( ϕ , c A ) A = A μ = ( ϕ , c A ) A=(A^(mu))=(phi,cA)A=\left(A^{\mu}\right)=(\phi, c \boldsymbol{A})A=(Aμ)=(ϕ,cA) is the 4 -vector potential with ϕ ϕ phi\phiϕ and A = A ( x ) A = A ( x ) A=A(x)\boldsymbol{A}=\boldsymbol{A}(x)A=A(x) being the electric scalar potential and the magnetic 3 -vector potential, respectively, and can be written as
(0.25) F = ( F μ ν ) = ( 0 E 1 E 2 E 3 E 1 0 c B 3 c B 2 E 2 c B 3 0 c B 1 E 3 c B 2 c B 1 0 ) (0.25) F = F μ ν = 0 E 1 E 2 E 3 E 1 0 c B 3 c B 2 E 2 c B 3 0 c B 1 E 3 c B 2 c B 1 0 {:(0.25)F=(F^(mu nu))=([0,-E^(1),-E^(2),-E^(3)],[E^(1),0,-cB^(3),cB^(2)],[E^(2),cB^(3),0,-cB^(1)],[E^(3),-cB^(2),cB^(1),0]):}F=\left(F^{\mu \nu}\right)=\left(\begin{array}{cccc} 0 & -E^{1} & -E^{2} & -E^{3} \tag{0.25}\\ E^{1} & 0 & -c B^{3} & c B^{2} \\ E^{2} & c B^{3} & 0 & -c B^{1} \\ E^{3} & -c B^{2} & c B^{1} & 0 \end{array}\right)(0.25)F=(Fμν)=(0E1E2E3E10cB3cB2E2cB30cB1E3cB2cB10)
which is a real antisymmetric matrix, i.e., F μ ν = F ν μ F μ ν = F ν μ F^(mu nu)=-F^(nu mu)F^{\mu \nu}=-F^{\nu \mu}Fμν=Fνμ, that combines both the electric and magnetic field strengths, i.e., E = ( E 1 , E 2 , E 3 ) E = E 1 , E 2 , E 3 E=(E^(1),E^(2),E^(3))\boldsymbol{E}=\left(E^{1}, E^{2}, E^{3}\right)E=(E1,E2,E3) and B = ( B 1 , B 2 , B 3 ) B = B 1 , B 2 , B 3 B=(B^(1),B^(2),B^(3))\boldsymbol{B}=\left(B^{1}, B^{2}, B^{3}\right)B=(B1,B2,B3). Using this tensor, Maxwell's equations can be written as
(0.26) μ F μ ν = j ν , μ F ν λ + ν F λ μ + λ F μ v = 0 (0.26) μ F μ ν = j ν , μ F ν λ + ν F λ μ + λ F μ v = 0 {:(0.26)del_(mu)F^(mu nu)=j^(nu)","quaddel^(mu)F^(nu lambda)+del^(nu)F^(lambda mu)+del^(lambda)F^(mu v)=0:}\begin{equation*} \partial_{\mu} F^{\mu \nu}=j^{\nu}, \quad \partial^{\mu} F^{\nu \lambda}+\partial^{\nu} F^{\lambda \mu}+\partial^{\lambda} F^{\mu v}=0 \tag{0.26} \end{equation*}(0.26)μFμν=jν,μFνλ+νFλμ+λFμv=0
where j = ( j μ ) = ( ρ , j ) j = j μ = ( ρ , j ) j=(j^(mu))=(rho,j)j=\left(j^{\mu}\right)=(\rho, \boldsymbol{j})j=(jμ)=(ρ,j) is the 4-current density with ρ = ρ ( x ) ρ = ρ ( x ) rho=rho(x)\rho=\rho(x)ρ=ρ(x) and j = j ( x ) j = j ( x ) j=j(x)\boldsymbol{j}=\boldsymbol{j}(x)j=j(x) being the charge density and the electric 3-current density, respectively. In addition, we have the two Lorentz invariants
(0.27) F μ ν F μ ν = 2 ( c 2 B 2 E 2 ) , ϵ μ ν λ ω F μ ν F λ ω = 8 c B E , (0.27) F μ ν F μ ν = 2 c 2 B 2 E 2 , ϵ μ ν λ ω F μ ν F λ ω = 8 c B E , {:(0.27)F^(mu nu)F_(mu nu)=2(c^(2)B^(2)-E^(2))","quadepsilon_(mu nu lambda omega)F^(mu nu)F^(lambda omega)=-8cB*E",":}\begin{equation*} F^{\mu \nu} F_{\mu \nu}=2\left(c^{2} \boldsymbol{B}^{2}-\boldsymbol{E}^{2}\right), \quad \epsilon_{\mu \nu \lambda \omega} F^{\mu \nu} F^{\lambda \omega}=-8 c \boldsymbol{B} \cdot \boldsymbol{E}, \tag{0.27} \end{equation*}(0.27)FμνFμν=2(c2B2E2),ϵμνλωFμνFλω=8cBE,
where ϵ μ ν λ ω ϵ μ ν λ ω epsilon_(mu nu lambda omega)\epsilon_{\mu \nu \lambda \omega}ϵμνλω is the Levi-Civita tensor with ϵ 0123 = ϵ 0123 = 1 ϵ 0123 = ϵ 0123 = 1 epsilon^(0123)=-epsilon_(0123)=1\epsilon^{0123}=-\epsilon_{0123}=1ϵ0123=ϵ0123=1. Maxwell's equations describe how sources (charges and currents) give rise to electric and magnetic fields. Assuming a moving test charge q q qqq with rest mass m m mmm and parametrizing the trajectory of the test charge as x = x ( s ) x = x ( s ) x=x(s)x=x(s)x=x(s), where s s sss is the proper time parameter, the Lorentz force law describes how the field strengths determine the trajectory of the test charge and is given by
(0.28) m c 2 x ¨ μ ( s ) = q x ˙ v ( s ) F μ v ( x ( s ) ) (0.28) m c 2 x ¨ μ ( s ) = q x ˙ v ( s ) F μ v ( x ( s ) ) {:(0.28)mc^(2)x^(¨)^(mu)(s)=qx^(˙)_(v)(s)F^(mu v)(x(s)):}\begin{equation*} m c^{2} \ddot{x}^{\mu}(s)=q \dot{x}_{v}(s) F^{\mu v}(x(s)) \tag{0.28} \end{equation*}(0.28)mc2x¨μ(s)=qx˙v(s)Fμv(x(s))
which is covariant under Lorentz transformations. The energy-momentum tensor T T TTT of the electromagnetic field is defined as
(0.29) T μ v = ϵ 0 F λ μ F λ v + ϵ 0 4 η μ v F λ ω F λ ω (0.29) T μ v = ϵ 0 F λ μ F λ v + ϵ 0 4 η μ v F λ ω F λ ω {:(0.29)T^(mu v)=epsilon_(0)F_(lambda)^(mu)F^(lambda v)+(epsilon_(0))/(4)eta^(mu v)F_(lambda omega)F^(lambda omega):}\begin{equation*} T^{\mu v}=\epsilon_{0} F_{\lambda}^{\mu} F^{\lambda v}+\frac{\epsilon_{0}}{4} \eta^{\mu v} F_{\lambda \omega} F^{\lambda \omega} \tag{0.29} \end{equation*}(0.29)Tμv=ϵ0FλμFλv+ϵ04ημvFλωFλω
where ϵ 0 ϵ 0 epsilon_(0)\epsilon_{0}ϵ0 is the electric constant (or permittivity of free space). It holds that T T TTT is symmetric, i.e., T μ ν = T ν μ T μ ν = T ν μ T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu}Tμν=Tνμ, and T μ μ = η μ ν T μ ν = 0 T μ μ = η μ ν T μ ν = 0 T_(mu)^(mu)=eta_(mu nu)T^(mu nu)=0T_{\mu}^{\mu}=\eta_{\mu \nu} T^{\mu \nu}=0Tμμ=ημνTμν=0. Furthermore, using Maxwell's equations, we obtain
(0.30) μ T μ ν = ϵ 0 j μ F μ ν = f v (0.30) μ T μ ν = ϵ 0 j μ F μ ν = f v {:(0.30)del_(mu)T^(mu nu)=epsilon_(0)j_(mu)F^(mu nu)=-f^(v):}\begin{equation*} \partial_{\mu} T^{\mu \nu}=\epsilon_{0} j_{\mu} F^{\mu \nu}=-f^{v} \tag{0.30} \end{equation*}(0.30)μTμν=ϵ0jμFμν=fv
where f = ( f μ ) = ( j E / c , ρ E + j × B ) f = f μ = ( j E / c , ρ E + j × B ) f=(f^(mu))=(j*E//c,rho E+j xx B)f=\left(f^{\mu}\right)=(\boldsymbol{j} \cdot \boldsymbol{E} / c, \rho \boldsymbol{E}+\boldsymbol{j} \times \boldsymbol{B})f=(fμ)=(jE/c,ρE+j×B) is the Lorentz force density generated by the 4 -current j j jjj. Without (external) sources, i.e., when j = 0 , T j = 0 , T j=0,Tj=0, Tj=0,T is conserved, i.e., μ T μ ν = 0 μ T μ ν = 0 del_(mu)T^(mu nu)=0\partial_{\mu} T^{\mu \nu}=0μTμν=0.

General Relativity

In a curved spacetime, the metric is defined as
(0.31) d s 2 = g μ ν d x μ d x ν (0.31) d s 2 = g μ ν d x μ d x ν {:(0.31)ds^(2)=g_(mu nu)dx^(mu)dx^(nu):}\begin{equation*} d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu} \tag{0.31} \end{equation*}(0.31)ds2=gμνdxμdxν
where the metric tensor and its inverse, i.e., the inverse metric tensor, are given by
(0.32) g = ( g μ ν ) , g 1 = ( g μ ν ) (0.32) g = g μ ν , g 1 = g μ ν {:(0.32)g=(g_(mu nu))","quadg^(-1)=(g^(mu nu)):}\begin{equation*} g=\left(g_{\mu \nu}\right), \quad g^{-1}=\left(g^{\mu \nu}\right) \tag{0.32} \end{equation*}(0.32)g=(gμν),g1=(gμν)
respectively. In the special case that the spacetime is flat (see Special Relativity), we obtain
(0.33) g μ ν = η μ ν , g μ ν = η μ ν (0.33) g μ ν = η μ ν , g μ ν = η μ ν {:(0.33)g_(mu nu)=eta_(mu nu)","quadg^(mu nu)=eta^(mu nu):}\begin{equation*} g_{\mu \nu}=\eta_{\mu \nu}, \quad g^{\mu \nu}=\eta^{\mu \nu} \tag{0.33} \end{equation*}(0.33)gμν=ημν,gμν=ημν
in Minkowski coordinates.
The covariant derivatives of a covariant vector A v A v A_(v)A_{v}Av and a contravariant vector A v A v A^(v)A^{v}Av are given by
(0.34) μ A ν = A v ; μ = μ A v Γ μ ν λ A λ , μ A ν = A ; μ v = μ A v + Γ μ λ v A λ (0.34) μ A ν = A v ; μ = μ A v Γ μ ν λ A λ , μ A ν = A ; μ v = μ A v + Γ μ λ v A λ {:(0.34)grad_(mu)A_(nu)=A_(v;mu)=del_(mu)A_(v)-Gamma_(mu nu)^(lambda)A_(lambda)","quadgrad_(mu)A^(nu)=A_(;mu)^(v)=del_(mu)A^(v)+Gamma_(mu lambda)^(v)A^(lambda):}\begin{equation*} \nabla_{\mu} A_{\nu}=A_{v ; \mu}=\partial_{\mu} A_{v}-\Gamma_{\mu \nu}^{\lambda} A_{\lambda}, \quad \nabla_{\mu} A^{\nu}=A_{; \mu}^{v}=\partial_{\mu} A^{v}+\Gamma_{\mu \lambda}^{v} A^{\lambda} \tag{0.34} \end{equation*}(0.34)μAν=Av;μ=μAvΓμνλAλ,μAν=A;μv=μAv+ΓμλvAλ
respectively. In particular, it holds that μ ν = Γ μ ν λ λ μ ν = Γ μ ν λ λ grad_(mu)del_(nu)=Gamma_(mu nu)^(lambda)del_(lambda)\nabla_{\mu} \partial_{\nu}=\Gamma_{\mu \nu}^{\lambda} \partial_{\lambda}μν=Γμνλλ, where the coefficients Γ μ ν λ Γ μ ν λ Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}Γμνλ are called the Christoffel symbols of the second kind. Given a metric g μ ν = g ν μ g μ ν = g ν μ g_(mu nu)=g_(nu mu)g_{\mu \nu}=g_{\nu \mu}gμν=gνμ, the Christoffel symbols of the Levi-Civita connection can be directly computed from
(0.35) Γ μ ν λ = 1 2 g λ ω ( μ g ν ω + ν g μ ω ω g μ ν ) (0.35) Γ μ ν λ = 1 2 g λ ω μ g ν ω + ν g μ ω ω g μ ν {:(0.35)Gamma_(mu nu)^(lambda)=(1)/(2)g^(lambda omega)(del_(mu)g_(nu omega)+del_(nu)g_(mu omega)-del_(omega)g_(mu nu)):}\begin{equation*} \Gamma_{\mu \nu}^{\lambda}=\frac{1}{2} g^{\lambda \omega}\left(\partial_{\mu} g_{\nu \omega}+\partial_{\nu} g_{\mu \omega}-\partial_{\omega} g_{\mu \nu}\right) \tag{0.35} \end{equation*}(0.35)Γμνλ=12gλω(μgνω+νgμωωgμν)
In addition, it holds that Γ μ ν λ = Γ ν μ λ Γ μ ν λ = Γ ν μ λ Gamma_(mu nu)^(lambda)=Gamma_(nu mu)^(lambda)\Gamma_{\mu \nu}^{\lambda}=\Gamma_{\nu \mu}^{\lambda}Γμνλ=Γνμλ, i.e., the Christoffel symbols are always symmetric with respect to the two lower indices.
The parallel transport equation for a vector A λ A λ A^(lambda)A^{\lambda}Aλ is given by
(0.36) x ˙ μ μ A λ = A ˙ λ + Γ μ ν λ x ˙ μ A v = 0 (0.36) x ˙ μ μ A λ = A ˙ λ + Γ μ ν λ x ˙ μ A v = 0 {:(0.36)x^(˙)^(mu)grad_(mu)A^(lambda)=A^(˙)^(lambda)+Gamma_(mu nu)^(lambda)x^(˙)^(mu)A^(v)=0:}\begin{equation*} \dot{x}^{\mu} \nabla_{\mu} A^{\lambda}=\dot{A}^{\lambda}+\Gamma_{\mu \nu}^{\lambda} \dot{x}^{\mu} A^{v}=0 \tag{0.36} \end{equation*}(0.36)x˙μμAλ=A˙λ+Γμνλx˙μAv=0
where the dot above ('‘') denotes differentiation with respect to the curve parameter s s sss. Furthermore, considering the Lagrangian density given by
(0.37) L = g μ ν x ˙ μ x ˙ ν (0.37) L = g μ ν x ˙ μ x ˙ ν {:(0.37)L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu):}\begin{equation*} \mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu} \tag{0.37} \end{equation*}(0.37)L=gμνx˙μx˙ν
and using the Euler-Lagrange equations, i.e.,
(0.38) L x μ d d s L x ˙ μ = 0 (0.38) L x μ d d s L x ˙ μ = 0 {:(0.38)(delL)/(delx^(mu))-(d)/(ds)(delL)/(delx^(˙)^(mu))=0:}\begin{equation*} \frac{\partial \mathcal{L}}{\partial x^{\mu}}-\frac{d}{d s} \frac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}=0 \tag{0.38} \end{equation*}(0.38)LxμddsLx˙μ=0
where μ = 0 , 1 , , n μ = 0 , 1 , , n mu=0,1,dots,n\mu=0,1, \ldots, nμ=0,1,,n, we obtain the geodesic equations as
(0.39) x ¨ λ + Γ μ ν λ x ˙ μ x ˙ ν = 0 (0.39) x ¨ λ + Γ μ ν λ x ˙ μ x ˙ ν = 0 {:(0.39)x^(¨)^(lambda)+Gamma_(mu nu)^(lambda)x^(˙)^(mu)x^(˙)^(nu)=0:}\begin{equation*} \ddot{x}^{\lambda}+\Gamma_{\mu \nu}^{\lambda} \dot{x}^{\mu} \dot{x}^{\nu}=0 \tag{0.39} \end{equation*}(0.39)x¨λ+Γμνλx˙μx˙ν=0
Given three vector fields X , Y X , Y X,YX, YX,Y, and Z Z ZZZ, the torsion T T TTT and the curvature R R RRR are defined as
(0.40) T ( X , Y ) = X Y Y X [ X , Y ] (0.41) R ( X , Y ) Z = [ X , Y ] Z [ X , Y ] Z (0.40) T ( X , Y ) = X Y Y X [ X , Y ] (0.41) R ( X , Y ) Z = X , Y Z [ X , Y ] Z {:[(0.40)T(X","Y)=grad_(X)Y-grad_(Y)X-[X","Y]],[(0.41)R(X","Y)Z=[grad_(X),grad_(Y)]Z-grad_([X,Y])Z]:}\begin{align*} T(X, Y) & =\nabla_{X} Y-\nabla_{Y} X-[X, Y] \tag{0.40}\\ R(X, Y) Z & =\left[\nabla_{X}, \nabla_{Y}\right] Z-\nabla_{[X, Y]} Z \tag{0.41} \end{align*}(0.40)T(X,Y)=XYYX[X,Y](0.41)R(X,Y)Z=[X,Y]Z[X,Y]Z
Both the torsion and curvature are tensors and are therefore linear in all of the arguments X , Y X , Y X,YX, YX,Y, and Z Z ZZZ, including when the arguments are multiplied by a scalar function f f fff, e.g.,
(0.42) T ( f X , Y ) = T ( X , f Y ) = f T ( X , Y ) , T ( X , Y + Z ) = T ( X , Y ) + T ( X , Z ) (0.42) T ( f X , Y ) = T ( X , f Y ) = f T ( X , Y ) , T ( X , Y + Z ) = T ( X , Y ) + T ( X , Z ) {:(0.42)T(fX","Y)=T(X","fY)=fT(X","Y)","quad T(X","Y+Z)=T(X","Y)+T(X","Z):}\begin{equation*} T(f X, Y)=T(X, f Y)=f T(X, Y), \quad T(X, Y+Z)=T(X, Y)+T(X, Z) \tag{0.42} \end{equation*}(0.42)T(fX,Y)=T(X,fY)=fT(X,Y),T(X,Y+Z)=T(X,Y)+T(X,Z)
Furthermore, the torsion and curvature tensors are antisymmetric in the arguments X X XXX and Y Y YYY as defined above
(0.43) T ( X , Y ) = T ( Y , X ) and R ( X , Y ) Z = R ( Y , X ) Z (0.43) T ( X , Y ) = T ( Y , X )  and  R ( X , Y ) Z = R ( Y , X ) Z {:(0.43)T(X","Y)=-T(Y","X)quad" and "quad R(X","Y)Z=-R(Y","X)Z:}\begin{equation*} T(X, Y)=-T(Y, X) \quad \text { and } \quad R(X, Y) Z=-R(Y, X) Z \tag{0.43} \end{equation*}(0.43)T(X,Y)=T(Y,X) and R(X,Y)Z=R(Y,X)Z
In local coordinates, we have
(0.44) T ( μ , ν ) λ λ = T μ ν λ λ , R ( μ , ν ) λ = R λ μ ν ω ω (0.44) T μ , ν λ λ = T μ ν λ λ , R μ , ν λ = R λ μ ν ω ω {:(0.44)T(del_(mu),del_(nu))^(lambda)del_(lambda)=T_(mu nu)^(lambda)del_(lambda)","quad R(del_(mu),del_(nu))del_(lambda)=R_(lambda mu nu)^(omega)del_(omega):}\begin{equation*} T\left(\partial_{\mu}, \partial_{\nu}\right)^{\lambda} \partial_{\lambda}=T_{\mu \nu}^{\lambda} \partial_{\lambda}, \quad R\left(\partial_{\mu}, \partial_{\nu}\right) \partial_{\lambda}=R_{\lambda \mu \nu}^{\omega} \partial_{\omega} \tag{0.44} \end{equation*}(0.44)T(μ,ν)λλ=Tμνλλ,R(μ,ν)λ=Rλμνωω
and the components of the torsion tensor and the Riemann curvature tensor may be computed as
(0.45) T μ ν λ = Γ μ ν λ Γ ν μ λ (0.46) R λ μ ν ω = μ Γ v λ ω ν Γ μ λ ω + Γ μ ρ ω Γ ν λ ρ Γ v ρ ω Γ μ λ ρ (0.45) T μ ν λ = Γ μ ν λ Γ ν μ λ (0.46) R λ μ ν ω = μ Γ v λ ω ν Γ μ λ ω + Γ μ ρ ω Γ ν λ ρ Γ v ρ ω Γ μ λ ρ {:[(0.45)T_(mu nu)^(lambda)=Gamma_(mu nu)^(lambda)-Gamma_(nu mu)^(lambda)],[(0.46)R_(lambda mu nu)^(omega)=del_(mu)Gamma_(v lambda)^(omega)-del_(nu)Gamma_(mu lambda)^(omega)+Gamma_(mu rho)^(omega)Gamma_(nu lambda)^(rho)-Gamma_(v rho)^(omega)Gamma_(mu lambda)^(rho)]:}\begin{align*} T_{\mu \nu}^{\lambda} & =\Gamma_{\mu \nu}^{\lambda}-\Gamma_{\nu \mu}^{\lambda} \tag{0.45}\\ R_{\lambda \mu \nu}^{\omega} & =\partial_{\mu} \Gamma_{v \lambda}^{\omega}-\partial_{\nu} \Gamma_{\mu \lambda}^{\omega}+\Gamma_{\mu \rho}^{\omega} \Gamma_{\nu \lambda}^{\rho}-\Gamma_{v \rho}^{\omega} \Gamma_{\mu \lambda}^{\rho} \tag{0.46} \end{align*}(0.45)Tμνλ=ΓμνλΓνμλ(0.46)Rλμνω=μΓvλωνΓμλω+ΓμρωΓνλρΓvρωΓμλρ
respectively. Note that the Levi-Civita connection is torsion free as Γ μ ν λ = Γ ν μ λ Γ μ ν λ = Γ ν μ λ Gamma_(mu nu)^(lambda)=Gamma_(nu mu)^(lambda)\Gamma_{\mu \nu}^{\lambda}=\Gamma_{\nu \mu}^{\lambda}Γμνλ=Γνμλ. For fixed μ μ mu\muμ and ν ν nu\nuν, we can write the Riemann curvature tensor in matrix form as
(0.47) R μ v = μ Γ v ν Γ μ + [ Γ μ , Γ v ] (0.47) R μ v = μ Γ v ν Γ μ + Γ μ , Γ v {:(0.47)R_(∙mu v)^(∙)=del_(mu)Gamma_(v∙)^(∙)-del_(nu)Gamma_(mu∙)^(∙)+[Gamma_(mu∙)^(∙),Gamma_(v∙)^(∙)]:}\begin{equation*} R_{\bullet \mu v}^{\bullet}=\partial_{\mu} \Gamma_{v \bullet}^{\bullet}-\partial_{\nu} \Gamma_{\mu \bullet}^{\bullet}+\left[\Gamma_{\mu \bullet}^{\bullet}, \Gamma_{v \bullet}^{\bullet}\right] \tag{0.47} \end{equation*}(0.47)Rμv=μΓvνΓμ+[Γμ,Γv]
Note that the Riemann curvature tensor is antisymmetric in μ μ mu\muμ and ν ν nu\nuν, i.e., R μ ν = R μ ν = R^(∙)_(∙mu nu)=R^{\bullet}{ }_{\bullet \mu \nu}=Rμν= R v μ R v μ -R^(∙)_(*v mu)-R^{\bullet}{ }_{\cdot v \mu}Rvμ, or in component form, R ω λ μ ν = R ω λ ν μ R ω λ μ ν = R ω λ ν μ R^(omega)_(lambda mu nu)=-R^(omega)_(lambda nu mu)R^{\omega}{ }_{\lambda \mu \nu}=-R^{\omega}{ }_{\lambda \nu \mu}Rωλμν=Rωλνμ. If the torsion vanishes, i.e., T = 0 T = 0 T=0T=0T=0, then we have the first Bianchi identity, i.e.,
(0.48) R ( X , Y ) Z + R ( Y , Z ) X + R ( Z , X ) Y = 0 (0.48) R ( X , Y ) Z + R ( Y , Z ) X + R ( Z , X ) Y = 0 {:(0.48)R(X","Y)Z+R(Y","Z)X+R(Z","X)Y=0:}\begin{equation*} R(X, Y) Z+R(Y, Z) X+R(Z, X) Y=0 \tag{0.48} \end{equation*}(0.48)R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0
or in component form, we have
(0.49) R λ μ ν ω + R μ ν λ ω + R ν λ μ ω = 0 . (0.49) R λ μ ν ω + R μ ν λ ω + R ν λ μ ω = 0 . {:(0.49)R_(lambda mu nu)^(omega)+R_(mu nu lambda)^(omega)+R_(nu lambda mu)^(omega)=0.:}\begin{equation*} R_{\lambda \mu \nu}^{\omega}+R_{\mu \nu \lambda}^{\omega}+R_{\nu \lambda \mu}^{\omega}=0 . \tag{0.49} \end{equation*}(0.49)Rλμνω+Rμνλω+Rνλμω=0.
Furthermore, we have the second Bianchi identity in matrix form, i.e.,
(0.50) μ R ν λ + [ Γ μ , R ν λ ] + ν R λ μ + [ Γ v , R λ μ ] + λ R μ ν + [ Γ λ , R μ ν ] = 0 (0.50) μ R ν λ + Γ μ , R ν λ + ν R λ μ + Γ v , R λ μ + λ R μ ν + Γ λ , R μ ν = 0 {:(0.50)del_(mu)R_(∙)^(∙)nu lambda+[Gamma_(mu∙)^(∙),R_(∙)^(∙)nu lambda]+del_(nu)R_(∙)^(∙)lambda mu+[Gamma_(v∙)^(∙),R_(∙)^(∙)lambda mu]+del_(lambda)R_(∙)^(∙)mu nu+[Gamma_(lambda_(∙),)^(∙)R_(∙)^(∙)mu nu]=0:}\begin{equation*} \partial_{\mu} R_{\bullet}^{\bullet} \nu \lambda+\left[\Gamma_{\mu \bullet}^{\bullet}, R_{\bullet}^{\bullet} \nu \lambda\right]+\partial_{\nu} R_{\bullet}^{\bullet} \lambda \mu+\left[\Gamma_{v \bullet}^{\bullet}, R_{\bullet}^{\bullet} \lambda \mu\right]+\partial_{\lambda} R_{\bullet}^{\bullet} \mu \nu+\left[\Gamma_{\lambda_{\bullet},}^{\bullet} R_{\bullet}^{\bullet} \mu \nu\right]=0 \tag{0.50} \end{equation*}(0.50)μRνλ+[Γμ,Rνλ]+νRλμ+[Γv,Rλμ]+λRμν+[Γλ,Rμν]=0
Using the Riemann curvature tensor, the Ricci tensor can be defined as
(0.51) R μ ν = R μ λ ν λ (0.51) R μ ν = R μ λ ν λ {:(0.51)R_(mu nu)=R_(mu lambda nu)^(lambda):}\begin{equation*} R_{\mu \nu}=R_{\mu \lambda \nu}^{\lambda} \tag{0.51} \end{equation*}(0.51)Rμν=Rμλνλ
which is symmetric, i.e., R μ ν = R ν μ R μ ν = R ν μ R_(mu nu)=R_(nu mu)R_{\mu \nu}=R_{\nu \mu}Rμν=Rνμ, and in turn, the Ricci scalar is defined as
(0.52) R = g μ v R μ ν = R μ μ (0.52) R = g μ v R μ ν = R μ μ {:(0.52)R=g^(mu v)R_(mu nu)=R_(mu)^(mu):}\begin{equation*} R=g^{\mu v} R_{\mu \nu}=R_{\mu}^{\mu} \tag{0.52} \end{equation*}(0.52)R=gμvRμν=Rμμ
Finally, the Einstein tensor is defined in terms of the Ricci tensor, the Ricci scalar, and the metric tensor as
(0.53) G μ ν = R μ ν 1 2 R g μ ν (0.53) G μ ν = R μ ν 1 2 R g μ ν {:(0.53)G_(mu nu)=R_(mu nu)-(1)/(2)Rg_(mu nu):}\begin{equation*} G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu} \tag{0.53} \end{equation*}(0.53)Gμν=Rμν12Rgμν
Note that it holds that the Einstein tensor is symmetric, i.e., G μ ν = G ν μ G μ ν = G ν μ G_(mu nu)=G_(nu mu)G_{\mu \nu}=G_{\nu \mu}Gμν=Gνμ, and conserved, i.e., its covariant divergence vanishes μ G μ ν = 0 μ G μ ν = 0 grad_(mu)G^(mu nu)=0\nabla_{\mu} G^{\mu \nu}=0μGμν=0. Under local coordinate transformations y = y ( x ) y = y ( x ) y=y(x)y=y(x)y=y(x), we have
(0.54) g μ ν ( y ) = x α y μ x β y v g α β ( x ) , (0.55) Γ μ ν λ ( y ) = x α y μ x β y v y λ x γ Γ α β γ ( x ) + y λ x γ 2 x γ y μ y v , (0.56) T μ ν λ ( y ) = y λ x γ x α y μ x β y v T α β γ ( x ) , (0.57) R λ μ ν ω ( y ) = y ω x δ x γ y λ x α y μ x β y ν R γ α β δ ( x ) . (0.54) g μ ν ( y ) = x α y μ x β y v g α β ( x ) , (0.55) Γ μ ν λ ( y ) = x α y μ x β y v y λ x γ Γ α β γ ( x ) + y λ x γ 2 x γ y μ y v , (0.56) T μ ν λ ( y ) = y λ x γ x α y μ x β y v T α β γ ( x ) , (0.57) R λ μ ν ω ( y ) = y ω x δ x γ y λ x α y μ x β y ν R γ α β δ ( x ) . {:[(0.54)g_(mu nu)^(')(y)=(delx^(alpha))/(dely^(mu))(delx^(beta))/(dely^(v))g_(alpha beta)(x)","],[(0.55)Gamma_(mu nu)^('lambda)(y)=(delx^(alpha))/(dely^(mu))(delx^(beta))/(dely^(v))(dely^(lambda))/(delx^(gamma))Gamma_(alpha beta)^(gamma)(x)+(dely^(lambda))/(delx^(gamma))(del^(2)x^(gamma))/(dely^(mu)dely^(v))","],[(0.56)T_(mu nu)^('lambda)(y)=(dely^(lambda))/(delx^(gamma))(delx^(alpha))/(dely^(mu))(delx^(beta))/(dely^(v))T_(alpha beta)^(gamma)(x)","],[(0.57)R_(lambda mu nu)^('omega)(y)=(dely^(omega))/(delx^(delta))(delx^(gamma))/(dely^(lambda))(delx^(alpha))/(dely^(mu))(delx^(beta))/(dely^(nu))R_(gamma alpha beta)^(delta)(x).]:}\begin{align*} g_{\mu \nu}^{\prime}(y) & =\frac{\partial x^{\alpha}}{\partial y^{\mu}} \frac{\partial x^{\beta}}{\partial y^{v}} g_{\alpha \beta}(x), \tag{0.54}\\ \Gamma_{\mu \nu}^{\prime \lambda}(y) & =\frac{\partial x^{\alpha}}{\partial y^{\mu}} \frac{\partial x^{\beta}}{\partial y^{v}} \frac{\partial y^{\lambda}}{\partial x^{\gamma}} \Gamma_{\alpha \beta}^{\gamma}(x)+\frac{\partial y^{\lambda}}{\partial x^{\gamma}} \frac{\partial^{2} x^{\gamma}}{\partial y^{\mu} \partial y^{v}}, \tag{0.55}\\ T_{\mu \nu}^{\prime \lambda}(y) & =\frac{\partial y^{\lambda}}{\partial x^{\gamma}} \frac{\partial x^{\alpha}}{\partial y^{\mu}} \frac{\partial x^{\beta}}{\partial y^{v}} T_{\alpha \beta}^{\gamma}(x), \tag{0.56}\\ R_{\lambda \mu \nu}^{\prime \omega}(y) & =\frac{\partial y^{\omega}}{\partial x^{\delta}} \frac{\partial x^{\gamma}}{\partial y^{\lambda}} \frac{\partial x^{\alpha}}{\partial y^{\mu}} \frac{\partial x^{\beta}}{\partial y^{\nu}} R_{\gamma \alpha \beta}^{\delta}(x) . \tag{0.57} \end{align*}(0.54)gμν(y)=xαyμxβyvgαβ(x),(0.55)Γμνλ(y)=xαyμxβyvyλxγΓαβγ(x)+yλxγ2xγyμyv,(0.56)Tμνλ(y)=yλxγxαyμxβyvTαβγ(x),(0.57)Rλμνω(y)=yωxδxγyλxαyμxβyνRγαβδ(x).
Symmetries of a spacetime metric are associated to so-called Killing vector fields. Consider a vector field X X XXX. By definition, X X XXX is a Killing vector field if
(0.58) μ X v + v X μ = 0 (0.58) μ X v + v X μ = 0 {:(0.58)grad_(mu)X_(v)+grad_(v)X_(mu)=0:}\begin{equation*} \nabla_{\mu} X_{v}+\nabla_{v} X_{\mu}=0 \tag{0.58} \end{equation*}(0.58)μXv+vXμ=0
for all indices μ μ mu\muμ and ν ν nu\nuν. Given a Killing vector field X μ X μ X^(mu)X^{\mu}Xμ and a geodesic described by coordinate functions x μ ( s ) x μ ( s ) x^(mu)(s)x^{\mu}(s)xμ(s), the quantity
(0.59) Q = x ˙ μ X μ = g μ ν x ˙ μ X v (0.59) Q = x ˙ μ X μ = g μ ν x ˙ μ X v {:(0.59)Q=x^(˙)^(mu)X_(mu)=g_(mu nu)x^(˙)^(mu)X^(v):}\begin{equation*} Q=\dot{x}^{\mu} X_{\mu}=g_{\mu \nu} \dot{x}^{\mu} X^{v} \tag{0.59} \end{equation*}(0.59)Q=x˙μXμ=gμνx˙μXv
is constant along the geodesic.
The dynamics of spacetime in vacuum are described in the Lagrange formalism using the Einstein-Hilbert action, namely
(0.60) S EH = M Pl 2 2 R | g ¯ | d 4 x (0.60) S EH = M Pl 2 2 R | g ¯ | d 4 x {:(0.60)S_(EH)=-(M_(Pl)^(2))/(2)int Rsqrt(| bar(g)|)d^(4)x:}\begin{equation*} \mathscr{S}_{\mathrm{EH}}=-\frac{M_{\mathrm{Pl}}^{2}}{2} \int R \sqrt{|\bar{g}|} d^{4} x \tag{0.60} \end{equation*}(0.60)SEH=MPl22R|g¯|d4x
where M Pl c 2 / 8 π G M Pl c 2 / 8 π G M_(Pl)-=c^(2)//sqrt(8pi G)M_{\mathrm{Pl}} \equiv c^{2} / \sqrt{8 \pi G}MPlc2/8πG is the Planck mass, R R RRR is the Ricci scalar, and g ¯ = det ( g ) g ¯ = det ( g ) bar(g)=det(g)\bar{g}=\operatorname{det}(g)g¯=det(g) is the determinant of the metric tensor. For the case of a spacetime not in vacuum, a matter contribution to the action is necessary
(0.61) S matter = L | g ¯ | d 4 x (0.61) S matter  = L | g ¯ | d 4 x {:(0.61)S_("matter ")=intLsqrt(| bar(g)|)d^(4)x:}\begin{equation*} \mathscr{S}_{\text {matter }}=\int \mathcal{L} \sqrt{|\bar{g}|} d^{4} x \tag{0.61} \end{equation*}(0.61)Smatter =L|g¯|d4x
where L L L\mathcal{L}L is the Lagrangian density of the matter contribution.
The Einstein gravitational field equations (or simply Einstein's equations) follow from the Euler-Lagrange equations for the action and are given by
(0.62) G μ ν = 8 π G c 4 T μ ν (0.62) G μ ν = 8 π G c 4 T μ ν {:(0.62)G_(mu nu)=(8pi G)/(c^(4))T_(mu nu):}\begin{equation*} G_{\mu \nu}=\frac{8 \pi G}{c^{4}} T_{\mu \nu} \tag{0.62} \end{equation*}(0.62)Gμν=8πGc4Tμν
where G G GGG is Newton's gravitational constant and T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν is the energy-momentum tensor (or the stress-energy tensor) that describes the distribution of energy in spacetime. The energy-momentum tensor is generally given by
(0.63) T μ ν = 2 | g ¯ | δ S matter δ g μ ν (0.63) T μ ν = 2 | g ¯ | δ S matter  δ g μ ν {:(0.63)T_(mu nu)=(2)/(sqrt(| bar(g)|))(deltaS_("matter "))/(deltag^(mu nu)):}\begin{equation*} T_{\mu \nu}=\frac{2}{\sqrt{|\bar{g}|}} \frac{\delta \mathscr{S}_{\text {matter }}}{\delta g^{\mu \nu}} \tag{0.63} \end{equation*}(0.63)Tμν=2|g¯|δSmatter δgμν
For example, an external electromagnetic field gives a contribution to T μ v T μ v T^(mu v)T^{\mu v}Tμv such that (see Special Relativity)
(0.64) T EM μ ν = ϵ 0 F λ μ F λ v + ϵ 0 4 g μ ν F λ ω F λ ω (0.64) T EM μ ν = ϵ 0 F λ μ F λ v + ϵ 0 4 g μ ν F λ ω F λ ω {:(0.64)T_(EM)^(mu nu)=epsilon_(0)F_(lambda)^(mu)F^(lambda v)+(epsilon_(0))/(4)g^(mu nu)F_(lambda omega)F^(lambda omega):}\begin{equation*} T_{\mathrm{EM}}^{\mu \nu}=\epsilon_{0} F_{\lambda}^{\mu} F^{\lambda v}+\frac{\epsilon_{0}}{4} g^{\mu \nu} F_{\lambda \omega} F^{\lambda \omega} \tag{0.64} \end{equation*}(0.64)TEMμν=ϵ0FλμFλv+ϵ04gμνFλωFλω
whereas a perfect fluid (characterized by a 4 -velocity u u uuu, a scalar density ρ 0 ρ 0 rho_(0)\rho_{0}ρ0, and a scalar pressure p p ppp ) gives
(0.65) T pf μ v = ( ρ 0 + p ) u μ u ν p g μ v (0.65) T pf μ v = ρ 0 + p u μ u ν p g μ v {:(0.65)T_(pf)^(mu v)=(rho_(0)+p)u^(mu)u^(nu)-pg^(mu v):}\begin{equation*} T_{\mathrm{pf}}^{\mu v}=\left(\rho_{0}+p\right) u^{\mu} u^{\nu}-p g^{\mu v} \tag{0.65} \end{equation*}(0.65)Tpfμv=(ρ0+p)uμuνpgμv
In vacuum, Einstein's equations reduce to G μ ν = 0 G μ ν = 0 G_(mu nu)=0G_{\mu \nu}=0Gμν=0.
In the Newtonian limit and the weak field approximation, i.e., g μ ν η μ ν + g μ ν η μ ν + g_(mu nu)≃eta_(mu nu)+g_{\mu \nu} \simeq \eta_{\mu \nu}+gμνημν+ h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν, where h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν is a small perturbation, the solutions to Einstein's equations are given by
(0.66) h 00 = h 11 = h 22 = h 33 = 2 c 2 Φ , h μ v = 0 μ v , (0.66) h 00 = h 11 = h 22 = h 33 = 2 c 2 Φ , h μ v = 0 μ v , {:(0.66)h_(00)=h_(11)=h_(22)=h_(33)=(2)/(c^(2))Phi","quadh_(mu v)=0quad AA mu!=v",":}\begin{equation*} h_{00}=h_{11}=h_{22}=h_{33}=\frac{2}{c^{2}} \Phi, \quad h_{\mu v}=0 \quad \forall \mu \neq v, \tag{0.66} \end{equation*}(0.66)h00=h11=h22=h33=2c2Φ,hμv=0μv,
where Φ Φ Phi\PhiΦ is the gravitational potential for the matter distribution ρ ρ rho\rhoρ and given by Φ = G M / r Φ = G M / r Phi=-GM//r\Phi=-G M / rΦ=GM/r, that is the solution to the Newtonian equation 2 Φ = 4 π G ρ 2 Φ = 4 π G ρ grad^(2)Phi=4pi G rho\nabla^{2} \Phi=4 \pi G \rho2Φ=4πGρ. Furthermore, the geodesic equations of motion become
(0.67) d 2 x i d t 2 = i Φ = i Φ (0.67) d 2 x i d t 2 = i Φ = i Φ {:(0.67)(d^(2)x^(i))/(dt^(2))=del^(i)Phi=-del_(i)Phi:}\begin{equation*} \frac{d^{2} x^{i}}{d t^{2}}=\partial^{i} \Phi=-\partial_{i} \Phi \tag{0.67} \end{equation*}(0.67)d2xidt2=iΦ=iΦ
The spherically symmetric vacuum solution to Einstein's equations is the Schwarzschild solution for which the Schwarzschild metric in spherical coordinates is given by
(0.68) d s 2 = g μ ν d x μ d x ν = ( 1 2 G M c 2 r ) c 2 d t 2 ( 1 2 G M c 2 r ) 1 d r 2 r 2 d Ω 2 (0.68) d s 2 = g μ ν d x μ d x ν = 1 2 G M c 2 r c 2 d t 2 1 2 G M c 2 r 1 d r 2 r 2 d Ω 2 {:(0.68)ds^(2)=g_(mu nu)dx^(mu)dx^(nu)=(1-(2GM)/(c^(2)r))c^(2)dt^(2)-(1-(2GM)/(c^(2)r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}=\left(1-\frac{2 G M}{c^{2} r}\right) c^{2} d t^{2}-\left(1-\frac{2 G M}{c^{2} r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{0.68} \end{equation*}(0.68)ds2=gμνdxμdxν=(12GMc2r)c2dt2(12GMc2r)1dr2r2dΩ2
where d Ω 2 d Ω 2 dOmega^(2)d \Omega^{2}dΩ2 describes the metric on a sphere, i.e., d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}dΩ2=dθ2+sin2θdϕ2. For large r r rrr, the Schwarzschild metric approaches the Minkowski metric. The particular value r = r 2 G M / c 2 r = r 2 G M / c 2 r=r_(**)-=2GM//c^(2)r=r_{*} \equiv 2 G M / c^{2}r=r2GM/c2 represents the Schwarzschild event horizon (or the Schwarzschild radius) and is a coordinate singularity, i.e., it can be removed by a change of coordinates. Such a coordinate change is given by Kruskal-Szekeres coordinates u , v , θ u , v , θ u,v,thetau, v, \thetau,v,θ, and ϕ ϕ phi\phiϕ, where θ θ theta\thetaθ and ϕ ϕ phi\phiϕ are the ordinary spherical coordinates on a unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2, the Kruskal-Szekeres metric is given by
(0.69) d s 2 = 16 μ 2 r e ( 2 μ r ) / ( 2 μ ) d u d v r 2 d Ω 2 , u v = ( 2 μ r ) e ( r 2 μ ) / ( 2 μ ) < 2 G M c 2 e (0.69) d s 2 = 16 μ 2 r e ( 2 μ r ) / ( 2 μ ) d u d v r 2 d Ω 2 , u v = ( 2 μ r ) e ( r 2 μ ) / ( 2 μ ) < 2 G M c 2 e {:(0.69)ds^(2)=(16mu^(2))/(r)e^((2mu-r)//(2mu))dudv-r^(2)dOmega^(2)","quad uv=(2mu-r)e^((r-2mu)//(2mu)) < (2GM)/(c^(2)e):}\begin{equation*} d s^{2}=\frac{16 \mu^{2}}{r} e^{(2 \mu-r) /(2 \mu)} d u d v-r^{2} d \Omega^{2}, \quad u v=(2 \mu-r) e^{(r-2 \mu) /(2 \mu)}<\frac{2 G M}{c^{2} e} \tag{0.69} \end{equation*}(0.69)ds2=16μ2re(2μr)/(2μ)dudvr2dΩ2,uv=(2μr)e(r2μ)/(2μ)<2GMc2e
where μ G M / c 2 μ G M / c 2 mu-=GM//c^(2)\mu \equiv G M / c^{2}μGM/c2.
For a static spacetime, the metric can be written on the form
(0.70) d s 2 = φ ( x ) 2 d t 2 g i j ( x ) d x i d x j (0.70) d s 2 = φ ( x ) 2 d t 2 g i j ( x ) d x i d x j {:(0.70)ds^(2)=varphi(x)^(2)dt^(2)-g_(ij)(x)dx^(i)dx^(j):}\begin{equation*} d s^{2}=\varphi(x)^{2} d t^{2}-g_{i j}(x) d x^{i} d x^{j} \tag{0.70} \end{equation*}(0.70)ds2=φ(x)2dt2gij(x)dxidxj
Given two static observers A A AAA and B B BBB in this spacetime, signals sent from A A AAA to B B BBB with frequencies f A f A f_(A)f_{A}fA and f B f B f_(B)f_{B}fB, respectively, will be redshifted according to
(0.71) z = f A f B 1 = φ ( x B ) φ ( x A ) 1 (0.71) z = f A f B 1 = φ x B φ x A 1 {:(0.71)z=(f_(A))/(f_(B))-1=(varphi(x_(B)))/(varphi(x_(A)))-1:}\begin{equation*} z=\frac{f_{A}}{f_{B}}-1=\frac{\varphi\left(x_{B}\right)}{\varphi\left(x_{A}\right)}-1 \tag{0.71} \end{equation*}(0.71)z=fAfB1=φ(xB)φ(xA)1
In particular, in the Schwarzschild spacetime, a signal sent from a static observer at r r rrr to an observer at infinity will be gravitationally redshifted according to
(0.72) z 1 1 2 G M c 2 r 1 G M c 2 r (0.72) z 1 1 2 G M c 2 r 1 G M c 2 r {:(0.72)z_(oo)-=(1)/(sqrt(1-(2GM)/(c^(2)r)))-1≃(GM)/(c^(2)r):}\begin{equation*} z_{\infty} \equiv \frac{1}{\sqrt{1-\frac{2 G M}{c^{2} r}}}-1 \simeq \frac{G M}{c^{2} r} \tag{0.72} \end{equation*}(0.72)z112GMc2r1GMc2r
where M M MMM is the mass of the gravitating body. More generally, the frequency f f fff of a light signal measured by an observer will be given by
(0.73) f = g μ ν U μ N ν (0.73) f = g μ ν U μ N ν {:(0.73)f=g_(mu nu)U^(mu)N^(nu):}\begin{equation*} f=g_{\mu \nu} U^{\mu} N^{\nu} \tag{0.73} \end{equation*}(0.73)f=gμνUμNν
where U U UUU is the 4 -velocity of the observer and N N NNN the 4 -frequency of the light signal, which is parallel transported along the worldline of the light signal.
In cosmology, the cosmological principles are encoded into the RobertsonWalker metric, which is given by
(0.74) d s 2 = c 2 d t 2 a ( t ) 2 ( d r 2 1 k r 2 + r 2 d Ω 2 ) (0.74) d s 2 = c 2 d t 2 a ( t ) 2 d r 2 1 k r 2 + r 2 d Ω 2 {:(0.74)ds^(2)=c^(2)dt^(2)-a(t)^(2)((dr^(2))/(1-kr^(2))+r^(2)dOmega^(2)):}\begin{equation*} d s^{2}=c^{2} d t^{2}-a(t)^{2}\left(\frac{d r^{2}}{1-k r^{2}}+r^{2} d \Omega^{2}\right) \tag{0.74} \end{equation*}(0.74)ds2=c2dt2a(t)2(dr21kr2+r2dΩ2)
where a ( t ) a ( t ) a(t)a(t)a(t) is some function of the universal time t t ttt and k k kkk is a constant. By a suitable coordinate transformation r λ r r λ r r|->lambda rr \mapsto \lambda rrλr, one can always choose λ λ lambda\lambdaλ such that k k kkk takes one of the three values k = 0 , ± 1 k = 0 , ± 1 k=0,+-1k=0, \pm 1k=0,±1. If k = 0 k = 0 k=0k=0k=0, then the spatial part for any fixed t t ttt becomes the Euclidean space R 3 R 3 R^(3)\mathbb{R}^{3}R3.
From Einstein's equations, the assumption of the Robertson-Walker metric, and the universe being filled by an ideal fluid, follow the two independent Friedmann equations, namely
(0.75) a ( t ) 2 + k c 2 a ( t ) 2 = 8 π G ρ + Λ c 2 3 (0.76) a ( t ) a ( t ) = 4 π G 3 ( ρ + 3 p c 2 ) + Λ c 2 3 (0.75) a ( t ) 2 + k c 2 a ( t ) 2 = 8 π G ρ + Λ c 2 3 (0.76) a ( t ) a ( t ) = 4 π G 3 ρ + 3 p c 2 + Λ c 2 3 {:[(0.75)(a^(')(t)^(2)+kc^(2))/(a(t)^(2))=(8pi G rho+Lambdac^(2))/(3)],[(0.76)(a^('')(t))/(a(t))=-(4pi G)/(3)(rho+(3p)/(c^(2)))+(Lambdac^(2))/(3)]:}\begin{align*} \frac{a^{\prime}(t)^{2}+k c^{2}}{a(t)^{2}} & =\frac{8 \pi G \rho+\Lambda c^{2}}{3} \tag{0.75}\\ \frac{a^{\prime \prime}(t)}{a(t)} & =-\frac{4 \pi G}{3}\left(\rho+\frac{3 p}{c^{2}}\right)+\frac{\Lambda c^{2}}{3} \tag{0.76} \end{align*}(0.75)a(t)2+kc2a(t)2=8πGρ+Λc23(0.76)a(t)a(t)=4πG3(ρ+3pc2)+Λc23
where the first equation is derived from the 00-component of Einstein's equations and the second equation is derived from the first one together with the trace of Einstein's equations. Here Λ Λ Lambda\LambdaΛ is the cosmological constant.

Conventions

In this book, the following conventions will mainly be used in the presentation of the problems and their corresponding solutions:
  • Units: We will mostly use units in which the speed of light in vacuum c c ccc has been set to c = 1 c = 1 c=1c=1c=1, these are usually known as natural units. In some problems, we have also set = 1 = 1 ℏ=1\hbar=1=1, if relevant. Normally, we do not use units in which Newton's gravitational constant G G GGG has been set to G = 1 G = 1 G=1G=1G=1. In some problems, it is useful to use SI units.
  • Vectors: In a four-dimensional spacetime, we will normally denote a 4-vector A A AAA by its contravariant components as follows
(0.77) A = ( A μ ) = ( A 0 , A 1 , A 2 , A 3 ) (0.77) A = A μ = A 0 , A 1 , A 2 , A 3 {:(0.77)A=(A^(mu))=(A^(0),A^(1),A^(2),A^(3)):}\begin{equation*} A=\left(A^{\mu}\right)=\left(A^{0}, A^{1}, A^{2}, A^{3}\right) \tag{0.77} \end{equation*}(0.77)A=(Aμ)=(A0,A1,A2,A3)
where A 0 A 0 A^(0)A^{0}A0 is the temporal component and A i ( i = 1 , 2 , 3 ) A i ( i = 1 , 2 , 3 ) A^(i)(i=1,2,3)A^{i}(i=1,2,3)Ai(i=1,2,3) are the spatial components, which is related to the 4 -vector expressed in its covariant components as follows
(0.78) ( A μ ) = ( A 0 , A 1 , A 2 , A 3 ) (0.78) A μ = A 0 , A 1 , A 2 , A 3 {:(0.78)(A_(mu))=(A_(0),A_(1),A_(2),A_(3)):}\begin{equation*} \left(A_{\mu}\right)=\left(A_{0}, A_{1}, A_{2}, A_{3}\right) \tag{0.78} \end{equation*}(0.78)(Aμ)=(A0,A1,A2,A3)
where it holds that A μ = g μ ν A ν A μ = g μ ν A ν A_(mu)=g_(mu nu)A^(nu)A_{\mu}=g_{\mu \nu} A^{\nu}Aμ=gμνAν with ( g μ ν ) g μ ν (g_(mu nu))\left(g_{\mu \nu}\right)(gμν) being the given metric tensor. In some textbooks, the convention that the temporal component of a 4 -vector is written as the last component of the vector is used, i.e., A = ( A 1 , A 2 , A 3 , A 4 ) A = A 1 , A 2 , A 3 , A 4 A=(A^(1),A^(2),A^(3),A^(4))A=\left(A^{1}, A^{2}, A^{3}, A^{4}\right)A=(A1,A2,A3,A4), whereas in other textbooks, the convention that the standard components of a 4 -vector are chosen as its covariant components might be used. We will not use these conventions.
  • Metrics: In four-dimensional spacetimes, we adopt the convention that the signature is +--- and, when relevant, place the temporal direction first and denote it by 0 . Therefore, in standard coordinates on Minkowski space, the metric tensor is ( η μ ν ) = diag ( 1 , 1 , 1 , 1 ) η μ ν = diag ( 1 , 1 , 1 , 1 ) (eta_(mu nu))=diag(1,-1,-1,-1)\left(\eta_{\mu \nu}\right)=\operatorname{diag}(1,-1,-1,-1)(ημν)=diag(1,1,1,1) (see Special Relativity) and its inverse is given by ( η μ ν ) = diag ( 1 , 1 , 1 , 1 ) η μ ν = diag ( 1 , 1 , 1 , 1 ) (eta^(mu nu))=diag(1,-1,-1,-1)\left(\eta^{\mu \nu}\right)=\operatorname{diag}(1,-1,-1,-1)(ημν)=diag(1,1,1,1). Thus, we have A μ = η μ ν A ν A μ = η μ ν A ν A_(mu)=eta_(mu nu)A^(nu)A_{\mu}=\eta_{\mu \nu} A^{\nu}Aμ=ημνAν, where A 0 = η 00 A 0 = A 0 A 0 = η 00 A 0 = A 0 A_(0)=eta_(00)A^(0)=A^(0)A_{0}=\eta_{00} A^{0}=A^{0}A0=η00A0=A0 and A i = η i i A i = A i A i = η i i A i = A i A_(i)=eta_(ii)A^(i)=-A^(i)A_{i}=\eta_{i i} A^{i}=-A^{i}Ai=ηiiAi=Ai (for fixed i = 1 , 2 , 3 i = 1 , 2 , 3 i=1,2,3i=1,2,3i=1,2,3 ). In a general coordinates, the metric is given by d s 2 = g μ ν d x μ d x ν d s 2 = g μ ν d x μ d x ν ds^(2)=g_(mu nu)dx^(mu)dx^(nu)d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}ds2=gμνdxμdxν (see General Relativity) and the metric tensor components represented in matrix form as g = ( g μ ν ) g = g μ ν g=(g_(mu nu))g=\left(g_{\mu \nu}\right)g=(gμν) from which its corresponding inverse components g 1 = ( g μ ν ) g 1 = g μ ν g^(-1)=(g^(mu nu))g^{-1}=\left(g^{\mu \nu}\right)g1=(gμν) can be computed. It must hold that g g 1 = g 1 g = 1 4 g g 1 = g 1 g = 1 4 gg^(-1)=g^(-1)g=1_(4)g g^{-1}=g^{-1} g=\mathbb{1}_{4}gg1=g1g=14, where 1 4 1 4 1_(4)\mathbb{1}_{4}14 is the 4 × 4 4 × 4 4xx44 \times 44×4 identity matrix.
  • Sign convention of the Levi-Civita pseudotensor: We define ϵ 0123 = + 1 ϵ 0123 = + 1 epsilon^(0123)=+1\epsilon^{0123}=+1ϵ0123=+1 (see Special Relativity), which means that with our convention for the Minkowski metric, we have ϵ 0123 = 1 ϵ 0123 = 1 epsilon_(0123)=-1\epsilon_{0123}=-1ϵ0123=1.
  • Partial derivatives: We will mostly denote covariant and contravariant partial derivatives as
(0.79) μ x μ , μ g μ ν ν , (0.79) μ x μ , μ g μ ν ν , {:(0.79)del_(mu)-=(del)/(delx^(mu))","quaddel^(mu)-=g^(mu nu)del_(nu)",":}\begin{equation*} \partial_{\mu} \equiv \frac{\partial}{\partial x^{\mu}}, \quad \partial^{\mu} \equiv g^{\mu \nu} \partial_{\nu}, \tag{0.79} \end{equation*}(0.79)μxμ,μgμνν,
where ( g μ ν ) g μ ν (g^(mu nu))\left(g^{\mu \nu}\right)(gμν) is the inverse metric tensor.
  • Covariant derivatives and Christoffel symbols: For covariant derivatives (see General Relativity), we will mostly use the notation μ μ grad_(mu)\nabla_{\mu}μ, but the notation D μ D μ D_(mu)D_{\mu}Dμ will sometimes be used. For Christoffel symbols of the second kind, we will only use the notation Γ μ ν λ Γ μ ν λ Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}Γμνλ (see General Relativity).
  • Sign convention of the Ricci tensor: The Ricci tensor may sometimes be defined as R μ ν = R μ ν λ λ R μ ν = R μ ν λ λ R_(mu nu)=R_(mu nu lambda)^(lambda)R_{\mu \nu}=R_{\mu \nu \lambda}^{\lambda}Rμν=Rμνλλ, which introduces a sign difference to our definition (see General Relativity) due to the antisymmetry of the Riemann curvature tensor as
(0.80) R μ ν λ λ = R μ λ ν λ . (0.80) R μ ν λ λ = R μ λ ν λ . {:(0.80)R_(mu nu lambda)^(lambda)=-R_(mu lambda nu)^(lambda).:}\begin{equation*} R_{\mu \nu \lambda}^{\lambda}=-R_{\mu \lambda \nu}^{\lambda} . \tag{0.80} \end{equation*}(0.80)Rμνλλ=Rμλνλ.
  • Sign convention of Einstein's equations: There is a sign convention in the definition of Einstein's equations, i.e., G μ ν = ± 8 π G T μ ν / c 4 G μ ν = ± 8 π G T μ ν / c 4 G_(mu nu)=+-8pi GT_(mu nu)//c^(4)G_{\mu \nu}= \pm 8 \pi G T_{\mu \nu} / c^{4}Gμν=±8πGTμν/c4, where we use the positive sign.
In general, it is important to keep in mind that different texts may use different conventions. In particular, the sign discrepancies in different expressions will often be due to differing sign conventions of the metric, the Levi-Civita pseudotensor, the Ricci tensor, and Einstein's equations.

Special Relativity Theory

1.1 Basics

Problem 1.1 a) In Figure 1.1, a spacetime diagram for an observer O O O\mathcal{O}O with an inertial frame is shown along with another observer O O O^(')\mathcal{O}^{\prime}O with another inertial frame. The 4 -vectors A , B , U A , B , U vec(A), vec(B), vec(U)\vec{A}, \vec{B}, \vec{U}A,B,U, and V V vec(V)\vec{V}V are drawn. Which of the following statements are true?
  1. In O O O\mathcal{O}O 's inertial frame, the scalar product between A A vec(A)\vec{A}A and B B vec(B)\vec{B}B is zero.
  2. In O O O^(')\mathcal{O}^{\prime}O 's inertial frame, the scalar product between A A vec(A)\vec{A}A and B B vec(B)\vec{B}B is zero.
  3. The scalar product between A A vec(A)\vec{A}A and B B vec(B)\vec{B}B is always nonzero.
  4. In O O O\mathcal{O}O 's inertial frame, the scalar product between U U vec(U)\vec{U}U and V V vec(V)\vec{V}V is zero.
  5. In O O O^(')\mathcal{O}^{\prime}O 's inertial frame, the scalar product between U U vec(U)\vec{U}U and V V vec(V)\vec{V}V is zero.
  6. The scalar product between U U vec(U)\vec{U}U and V V vec(V)\vec{V}V is always nonzero.
    b) Which of the 4 -vectors in a) could be proportional to a 4 -velocity? Explain why.

Problem 1.2 Show that

a) every 4 -vector (i.e., vector in Minkowski space) that is orthogonal to a timelike 4 -vector is spacelike.
b) the sum of two future directed time-like 4 -vectors is another future directed timelike 4-vector.
c) every space-like 4 -vector can be written as the difference between two futuredirected lightlike 4 -vectors.
d) the inner product of two future-directed timelike 4-vectors is positive.
Problem 1.3 In a particular inertial frame, two observers have the 3-velocities v 1 v 1 v_(1)\boldsymbol{v}_{1}v1 and v 2 v 2 v_(2)\boldsymbol{v}_{2}v2, respectively. Find an expression for the gamma factor of observer 2 in the rest frame of observer 1 in terms of these velocities.
Problem 1.4 a) Can a rest frame be chosen for a photon? Explain why!
  1. always
  2. sometimes
  3. never
Figure 1.1 Spacetime diagram for observers O O O\mathcal{O}O and O O O^(')\mathcal{O}^{\prime}O.
b) Can a rest frame be chosen for the center of momentum for a system of two photons? Explain why!
  1. always
  2. sometimes
  3. never

1.2 Length Contraction, Time Dilation, and Spacetime Diagrams

Problem 1.5 a) State, explain, and derive the formula for length contraction in special relativity.
b) State, explain, and derive the formula for time dilation in special relativity.
Problem 1.6 A rod with length of 1 m is inclined 45 45 45^(@)45^{\circ}45 in the x y x y xyx yxy-plane with respect to the x x xxx-axis. An observer with the speed 2 / 3 c 2 / 3 c sqrt(2//3)c\sqrt{2 / 3} c2/3c approaches the rod in the positive direction along the x x xxx-axis. How long does the observer measure the rod to be and at which angle does (s)he observe it to be inclined relative to its x x xxx-axis?
Problem 1.7 When the primary cosmic rays hit the atmosphere, muons are created at an altitude between 10 km and 20 km . A muon in the laboratory lives on average the time τ 0 = 2.2 10 6 s τ 0 = 2.2 10 6 s tau_(0)=2.2*10^(-6)s\tau_{0}=2.2 \cdot 10^{-6} \mathrm{~s}τ0=2.2106 s before it decays into an electron (or a positron) and two neutrinos. Even though a muon can only move τ 0 c 660 m τ 0 c 660 m tau_(0)c~~660m\tau_{0} c \approx 660 \mathrm{~m}τ0c660 m under the time τ 0 τ 0 tau_(0)\tau_{0}τ0, a large fraction of the muons will reach the surface of the Earth. How can this be explained? Make a numerical computation for a muon that moves with velocity 0.999 c 0.999 c 0.999 c0.999 c0.999c.
Problem 1.8 An express train passes a station with velocity v v vvv. A measurement of the length of the train can be performed in the following different ways:
a) A "continuum" of linesmen is ordered to align along the track. The two men that see the front or the end of the train pass in front of them when their watches show 12:30 make a mark where they stand. The distance L a L a L_(a)L_{a}La between the marks is measured.
b) One conductor goes to the front of the train and another goes to the end. When the watches of the conductors show 12 : 15 12 : 15 12:1512: 1512:15, they quickly drive a nail into the track. The linesmen measure the distance L b L b L_(b)L_{b}Lb between the nails.
c) The stationmaster inspects the receding train through a pair of binoculars. Through the binoculars the stationmaster sees the front of the train to be at the semaphore A A AAA at the same time as its end is at the railway point B B BBB. The linesmen measure the distance L c L c L_(c)L_{c}Lc between A A AAA and B B BBB.
d) The stationmaster uses a radar to measure the length of the train. The arrival times of the radar pulses reflected from the front and end of the receding train are t 1 t 1 t_(1)t_{1}t1 and t 2 t 2 t_(2)t_{2}t2, respectively. The distance L d = ( t 1 t 2 ) c / 2 L d = t 1 t 2 c / 2 L_(d)=(t_(1)-t_(2))c//2L_{d}=\left(t_{1}-t_{2}\right) c / 2Ld=(t1t2)c/2 is a measure of the length of the train.
Express L a , L b , L c L a , L b , L c L_(a),L_(b),L_(c)L_{a}, L_{b}, L_{c}La,Lb,Lc, and L d L d L_(d)L_{d}Ld in terms of L 0 L 0 L_(0)L_{0}L0, the rest length of the train.
Problem 1.9 A hitchhiker in the Milky Way sits waiting on a small asteroid when a formidably long express space cruiser passes very close to the asteroid. Just as the rear end is opposite to the hitchhiker, (s)he sees lanterns in the front and in the rear end of the cruiser go on simultaneously. Actually, the rear watchman also saw them go on, but according to his hydrogen maser wristwatch he measured a small time difference of 4 10 9 s 4 10 9 s 4*10^(-9)s4 \cdot 10^{-9} \mathrm{~s}4109 s between the lightening of the forward and rear lanterns. From the type indication on the cruiser - X2000 - our hitchhiker realized that its length was 2 10 3 m 2 10 3 m 2*10^(3)m2 \cdot 10^{3} \mathrm{~m}2103 m. Had they known what you know, they could have calculated the speed of the cruiser. What was it, according to Einstein's special theory of relativity?
Problem 1.10 Two lamps, which are separated by the distance \ell in an inertial coordinate system K K KKK, are switched on simultaneously (in K K KKK ). In another inertial coordinate system K K K^(')K^{\prime}K, an observer measures the distance between the lamps to be ℓ^(')\ell^{\prime} and observes the lamps go on with the time difference τ τ tau\tauτ. Express \ell in terms of ℓ^(')\ell^{\prime} and τ τ tau\tauτ. Assuming that the inertial coordinate system K K K^(')K^{\prime}K is moving along the axis connecting the two lamps, also find the expression for the relative velocity v v vvv between the two inertial coordinate systems.
Problem 1.11 A rod of length \ell lies in the x z x z xzx zxz-plane of a coordinate system. If the angle between the rod and the x x xxx-axis is θ θ theta\thetaθ, calculate the length of the rod as seen by an observer moving with velocity v v vvv along the x x xxx-axis.
Problem 1.12 Two events A A AAA and B B BBB with coordinates x A x A x_(A)x_{A}xA and x B x B x_(B)x_{B}xB are simultaneous for an observer K K KKK with rest frame S S SSS. Another observer, K K K^(')K^{\prime}K, moving with velocity -u along the x x xxx-axis of S S SSS measures these events to not be simultaneous, but such that B B BBB is earlier than A A AAA by the amount Δ t Δ t Deltat^(')\Delta t^{\prime}Δt. What is the distance L L LLL between the events A A AAA and B B BBB expressed in the frame of K K KKK if it is L L L^(')L^{\prime}L in the rest frame of K K KKK ?
Problem 1.13 An observer S S SSS with rest frame K K KKK observes two events x α x α x_(alpha)x_{\alpha}xα and x β x β x_(beta)x_{\beta}xβ. The α α alpha\alphaα event takes place at the origin and the β β beta\betaβ event 2 years later at a distance of 10 light years (ly) forward along the x 1 x 1 x^(1)x^{1}x1-axis. Another observer S S S^(')S^{\prime}S with rest frame K K K^(')K^{\prime}K moves with velocity v v vvv along the x 1 x 1 x^(1)x^{1}x1-axis of K K KKK, passing S S SSS at the origin. The observer S S S^(')S^{\prime}S instead observes the β β beta\betaβ event 1 year later than the α α alpha\alphaα event.
a) How far away does S S S^(')S^{\prime}S find the β β beta\betaβ event?
b) What is the relative velocity between S S SSS and S S S^(')S^{\prime}S ?
Problem 1.14 The ratio R ( μ / e ) R ( μ / e ) R(mu//e)R(\mu / e)R(μ/e) of muon neutrinos to electron neutrinos measured at ground level from the cosmic radiation is R ( μ / e ) = 2 R ( μ / e ) = 2 R(mu//e)=2R(\mu / e)=2R(μ/e)=2 at low energies. These neutrinos come from the decay of pions, created by the primary cosmic radiation, which consists mostly of protons. The relevant reaction chain can be written in simplified form as follows
π μ + v μ μ e + v μ + v e . π μ + v μ μ e + v μ + v e . {:[pi longrightarrow mu+v_(mu)],[mu longrightarrow e+v_(mu)+v_(e).]:}\begin{aligned} & \pi \longrightarrow \mu+v_{\mu} \\ & \mu \longrightarrow e+v_{\mu}+v_{e} . \end{aligned}πμ+vμμe+vμ+ve.
As we can see there are two muon neutrinos v μ v μ v_(mu)v_{\mu}vμ produced for every electron neutrino v e v e v_(e)v_{e}ve. When the energy of the muon neutrinos, and therefore the muons, is high enough this ratio goes up, since the muons hit the Earth before they decay, and no electron neutrinos are produced. In the muon's rest frame, the muon lifetime is τ 0 = 2.2 μ s τ 0 = 2.2 μ s tau_(0)=2.2 mus\tau_{0}=2.2 \mu \mathrm{~s}τ0=2.2μ s. The speed of light is 3 10 8 m / s 3 10 8 m / s 3*10^(8)m//s3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}3108 m/s. What is the smallest energy of the muons that hit the ground before they decay substantially if they are produced at an altitude of 10 km above ground? The rest mass of the muon is 106 MeV .
Problem 1.15 A circular accelerator has a radius of 50 m . How many turns can a muon take on average in this ring before it decays if its energy is kept constant at 1 GeV ? The average lifetime of the muon in its rest frame is 2.2 μ s 2.2 μ s 2.2 mus2.2 \mu \mathrm{~s}2.2μ s and the muon mass is 106 MeV .
Problem 1.16 Consider a triangle at rest in the inertial system K K KKK with sides of length a = 3 , b = 4 a = 3 , b = 4 a=3ℓ,b=4ℓa=3 \ell, b=4 \ella=3,b=4, and c = 5 c = 5 c=5ℓc=5 \ellc=5 in K K KKK.
a) Compute the lengths of the sides and the area of this triangle as measured in an inertial frame K K K^(')K^{\prime}K moving with constant velocity v v vvv parallel to the a a aaa-side of the triangle.
b) Same as in a), but now the observer K K K^(')K^{\prime}K moves parallel to the c c ccc-side of the triangle.
Problem 1.17 Consider a pole of proper length L L LLL moving along the x x xxx-axis in the negative direction with a constant velocity so that the pole is parallel to the x x xxx-axis (see Figure 1.2). At a fixed time, an observer at rest at the spatial origin sees (the optical effect is referred to) the front of the pole at an angle π / 3 π / 3 pi//3\pi / 3π/3 with the x x xxx-axis, a mark on the pole at an angle π / 4 π / 4 pi//4\pi / 4π/4, and the end of the pole at an angle π / 6 π / 6 pi//6\pi / 6π/6. What is the quotient r r rrr between the distance from the front of the pole to the mark and the full length of the pole?
Problem 1.18 Two spaceships, which are initially at rest in some common rest frame, are connected by a straight tensionless string. At time t = 0 t = 0 t=0t=0t=0 in this frame, both spaceships start to accelerate in the same direction, in the direction of the string, such that their separation is constant in the initial rest frame. Both spaceships agree to stop accelerating once a predetermined time t 0 t 0 t_(0)t_{0}t0 has passed in the initial rest frame.
a) Does the string break, i.e., does the distance between the two spaceships increase in the new rest frame of the spaceships?
b) If the distance between the spaceships is originally 40 km , t 0 = 30 s 40 km , t 0 = 30 s 40km,t_(0)=30s40 \mathrm{~km}, t_{0}=30 \mathrm{~s}40 km,t0=30 s, and the spaceships have constant acceleration of 1 / 50 c / s 1 / 50 c / s 1//50c//s1 / 50 \mathrm{c} / \mathrm{s}1/50c/s in the initial rest frame, what is the
Figure 1.2 The pole is moving in the negative direction of the x x xxx-axis with a constant velocity v v vvv. The z z zzz-direction is neglected.
distance between the two spaceships in the frame of the leading spaceship after the engines are turned off?
Problem 1.19 Professor A. Einstein is traveling in a train on a rainy night. He is situated in the exact middle of the train, and suddenly lightning strikes right next to him. The train has reflectors in the rear and front, and since the reflections from rear and front reach him at the same time, he falls into slumber convinced that the reflections happened at the same time and that the speed of light is the same in both directions. What he did not see was that Professor W. Wolf was standing on the ground, also next to the lightning strike, observing the events. Draw spacetime diagrams showing how the light signals travel in each of the professors' rest frames. Use these to answer (including motivation) the following
a) Does Professor Wolf see the light reflections reaching Professor Einstein at the same time?
b) Would Professor Wolf agree with Professor Einstein that the light signals were reflected at the same time?
c) Do the reflections reach Professor Wolf at the same time?
Problem 1.20 Two rockets with rest lengths L L LLL and 2 L 2 L 2L2 L2L, respectively, move with constant velocities on an interstellar highway. Since the velocities are different, the rockets will pass each other. Call the event when the front of the faster rocket reaches the slower rocket A A AAA and the event when the end of the faster rocket reaches the front of the slower rocket B B BBB (see Figure 1.3). In each rocket there is an observer. Draw one or more spacetime diagrams describing the events, and use it/them to deduce which observer will consider time between A A AAA and B B BBB to be larger (the observer in the short rocket or the observer in the long rocket).
Problem 1.21 Muons created by cosmic rays hitting the atmosphere have a lifetime of 2.2 10 6 s 2.2 10 6 s 2.2*10^(-6)s2.2 \cdot 10^{-6} \mathrm{~s}2.2106 s. If the muons are created at a height of 10 km , the time to reach the surface of the Earth (measured in the rest frame of the Earth) is at least 10 km / c 3 10 5 s 10 km / c 3 10 5 s 10km//c≃3*10^(-5)s10 \mathrm{~km} / c \simeq 3 \cdot 10^{-5} \mathrm{~s}10 km/c3105 s, yet a large fraction of the muons can be measured at sea level.
Figure 1.3 Two rockets with rest-lengths L L LLL and 2 L 2 L 2L2 L2L, respectively. Part (a) of the figure shows the event A A AAA, whereas part (b) shows the event B B BBB.
Figure 1.4 Part (a) of the figure shows the dimensions of a guillotine blade in its rest frame, whereas part (b) shows the event of the Scarlet Pimpernel riding by on his horse at velocity v v vvv ( S S S^(')S^{\prime}S is the rest frame of the Scarlet Pimpernel and his horse) and the guillotine blade falls at velocity u u uuu in its own rest frame.
Explain qualitatively why this occurs by describing the situation using spacetime diagrams.
Problem 1.22 During the French Revolution, guillotines with a slanted blade were used to decapitate nobility. The guillotine blade at rest has the dimensions shown in Figure 1.4. Eager to save the nobility, the Scarlet Pimpernel rides by on his horse at velocity v v vvv. How fast does he have to ride in order for the guillotine blade to be horizontal in his rest frame S S S^(')S^{\prime}S if it falls at velocity u u uuu in the guillotine rest frame?
Problem 1.23 In an inertial frame S S SSS, two lights located on the positive x x xxx-axis are moving in the negative x x xxx-direction at speed v v vvv. An observer placed in the origin of S S SSS looks at the light signals coming from the lights. What is the distance between the seen positions of the lights if their separation in their common rest frame is 0 0 ℓ_(0)\ell_{0}0 ?
Note: The problem is asking for the separation as seen by the observer, not the actual distance between the lights at a given time.

1.3 Lorentz Transformations and Geometry of Minkowski Space

Problem 1.24 Verify directly from the form of the Lorentz transformation representing a boost in the x x xxx-direction that any object traveling at speed c c ccc in an inertial frame S S SSS travels at speed c c ccc in the boosted frame.
Problem 1.25 A train passes a station just after sunset. The length of the train is L L LLL. In the front and in the rear, it has two lanterns. The lanterns are turned on simultaneously in the train's rest frame. A stationman observes the train pass with velocity v v vvv. Does the stationman see the lanterns go on simultaneously? If not, what is the time difference between the turning on of the two lanterns for the stationman, expressed in terms of L L LLL and v v vvv ?
Problem 1.26 An observer O O OOO on a train of length L L LLL and velocity v v vvv relative to the ground is standing at a distance x L ( 0 x 1 ) x L ( 0 x 1 ) xL(0 <= x <= 1)x L(0 \leq x \leq 1)xL(0x1) from the front A A AAA of the train. When the light from the lamps at A A AAA and B B BBB, at the rear, reach him/her simultaneously, (s)he can calculate at which times t 1 ( A ) t 1 ( A ) t_(1)(A)t_{1}(A)t1(A) and t 2 ( B ) t 2 ( B ) t_(2)(B)t_{2}(B)t2(B) they turned on. Another observer O O O^(')O^{\prime}O on the ground can also determine these two times t 1 t 1 t_(1)^(')t_{1}^{\prime}t1 and t 2 t 2 t_(2)^(')t_{2}^{\prime}t2 in his rest frame, where the light reaches him as O O OOO just passes him. If (s)he then finds that t 1 = t 2 t 1 = t 2 t_(1)^(')=t_(2)^(')t_{1}^{\prime}=t_{2}^{\prime}t1=t2, it turns out that the velocity v v vvv of the train can be expressed as a rather simple function of x x xxx. Find this function and show that if v = 0 v = 0 v=0v=0v=0, then x = 1 / 2 x = 1 / 2 x=1//2x=1 / 2x=1/2.
Problem 1.27 A particle of mass m m mmm and energy E E EEE falls from zenith to the Earth along the z z zzz-axis in the rest frame of observer K K KKK. Another observer, K K K^(')K^{\prime}K, moves with velocity v v vvv along the positive x x xxx-axis of K K KKK and will observe the particle to approach K K K^(')K^{\prime}K with an angle θ θ theta\thetaθ relative to the z z z^(')z^{\prime}z-axis.
a) Calculate the angle θ θ theta\thetaθ expressed in terms of the velocity u u uuu of the particle and the velocity v v vvv of K K K^(')K^{\prime}K.
b) Based on the result of a) give a description of how the starry sky would look like for a space cruiser moving with high speed in our galaxy.
Problem 1.28 Consider a particle with 4-velocity V = γ ( v ) ( c , v , 0 , 0 ) V = γ v c , v , 0 , 0 V=gamma(v^('))(c,v^('),0,0)V=\gamma\left(v^{\prime}\right)\left(c, v^{\prime}, 0,0\right)V=γ(v)(c,v,0,0). By making a Lorentz transformation with velocity v v -v-vv along the x 1 x 1 x^(1)x^{1}x1-axis, show that you can obtain the formula for relativistic addition of velocities, by expressing the velocity v v v^('')v^{\prime \prime}v of the particle in the new system in terms of the velocity v v v^(')v^{\prime}v in the old system and the velocity v v vvv of the motion of the observer.
Problem 1.29 Consider an equilateral triangle with sides of length \ell, which is at rest in the inertial coordinate system K K KKK. Assume that one of the sides in the triangle is parallel to the x 1 x 1 x^(1)x^{1}x1-axis of K K KKK. In an inertial coordinate system K K K^(')K^{\prime}K moving relative to K K KKK with velocity v v vvv along the positive x 1 x 1 x^(1)x^{1}x1-axis of K K KKK, an observer measures the lengths of the sides and angles in the triangle. What expressions in \ell and v v vvv for the lengths and angles does the observer find?
Problem 1.30 An observer K K K^(')K^{\prime}K is moving with constant speed v v vvv along the positive x 1 x 1 x^(1)x^{1}x1-axis of an observer K K KKK. A thin rod is parallel to the x 1 x 1 x^('1)x^{\prime 1}x1-axis and moving in the direction of the positive x 2 x 2 x^('2)x^{\prime 2}x2-axis with relative velocity u u uuu. Show that according to the observer K K KKK the rod forms an angle ϕ ϕ phi\phiϕ with the x 1 x 1 x^(1)x^{1}x1-axis, with
(1.1) tan ϕ = u v / c 2 1 v 2 / c 2 (1.1) tan ϕ = u v / c 2 1 v 2 / c 2 {:(1.1)tan phi=-(uv//c^(2))/(sqrt(1-v^(2)//c^(2))):}\begin{equation*} \tan \phi=-\frac{u v / c^{2}}{\sqrt{1-v^{2} / c^{2}}} \tag{1.1} \end{equation*}(1.1)tanϕ=uv/c21v2/c2
Problem 1.31 A cylinder is rotating around its axis with angular velocity ω ω omega\omegaω (rad/s) in an inertial system. A straight line is drawn along the length of the cylinder. Show that the observer in an inertial system, which moves with velocity v v vvv parallel to the direction of the cylinder axis, will measure the line as twisted around the cylinder. Determine the twist angle per unit length.
Problem 1.32 A fast train (velocity v v vvv ) is passing a station during the night. As the train passes the station, all compartment lights are turned on simultaneously with respect to the rest frame of the train. Relative to an observer standing at the station, the lights seem to be turned on at various times. Compute the velocity u u uuu of the line separating the illuminated and unilluminated parts of the train in the station rest frame.
Problem 1.33 A planet is moving along a circular orbit (radius R R RRR and angular velocity ω ω omega\omegaω ) around a star. A space ship is passing by the star, orthogonal with respect to the plane of motion of the planet, with velocity v v vvv. Compute the orbit of the planet in the rest frame coordinates of the space ship.
Problem 1.34 An observer B B BBB is moving with constant velocity v v vvv along the positive x 1 x 1 x^(1)x^{1}x1-axis in the rest frame K K KKK of an observer A A AAA. An observer C C CCC is moving with constant velocity v v v^(')v^{\prime}v along the positive x 2 2 x 2 2 x^('2)^(2){x^{\prime 2}}^{2}x22-axis in the rest frame K K K^(')K^{\prime}K of the observer B B BBB. Compute the absolute value of the relative velocity of C C CCC with respect to A A AAA. What is the time interval Δ t Δ t Delta t\Delta tΔt between two events E 1 E 1 E_(1)E_{1}E1 and E 2 E 2 E_(2)E_{2}E2 that occur at the same spatial point with time difference Δ t Δ t Deltat^('')\Delta t^{\prime \prime}Δt in the rest frame K K K^('')K^{\prime \prime}K of observer C C CCC.
Hint: It is sufficient to compute the time coordinate x 0 x 0 x^(''0)x^{\prime \prime 0}x0 of C C CCC as a function of the coordinates x μ x μ x^(mu)x^{\mu}xμ of A A AAA.
Problem 1.35 Let x x xxx be a lightlike vector in Minkowski space. Show that
(1.2) u = N ( x 0 + x 3 x 1 + i x 2 ) (1.2) u = N ( x 0 + x 3 x 1 + i x 2 ) {:(1.2)u=N((x^(0)+x^(3))/(x^(1)+ix^(2))):}\begin{equation*} u=N\binom{x^{0}+x^{3}}{x^{1}+i x^{2}} \tag{1.2} \end{equation*}(1.2)u=N(x0+x3x1+ix2)
where N N NNN is a real normalization factor, u u uuu is a spinor that satisfies X u u X u u X prop uu^(**)X \propto u u^{*}Xuu, where X X XXX is a complex 2 × 2 2 × 2 2xx22 \times 22×2 matrix, so that det X = det ( u u ) = 0 det X = det u u = 0 det X=det(uu^(**))=0\operatorname{det} X=\operatorname{det}\left(u u^{*}\right)=0detX=det(uu)=0. Normalize this spinor by the requirement that tr X = 2 x 0 tr X = 2 x 0 tr X=2x^(0)\operatorname{tr} X=2 x^{0}trX=2x0.
A Lorentz transformation along the 3-axis is given by
(1.3) a ( v ) = ( e θ / 2 0 0 e θ / 2 ) , (1.3) a ( v ) = e θ / 2 0 0 e θ / 2 , {:(1.3)a(v)=([e^(-theta//2),0],[0,e^(theta//2)])",":}a(v)=\left(\begin{array}{cc} e^{-\theta / 2} & 0 \tag{1.3}\\ 0 & e^{\theta / 2} \end{array}\right),(1.3)a(v)=(eθ/200eθ/2),
where tanh θ = v / c tanh θ = v / c tanh theta=v//c\tanh \theta=v / ctanhθ=v/c. Show explicitly that this transformation satisfies
(1.4) a ( v ) u = u ( L ( a ( v ) ) x ) , (1.4) a ( v ) u = u ( L ( a ( v ) ) x ) , {:(1.4)a(v)u=u(L(a(v))x)",":}\begin{equation*} a(v) u=u(L(a(v)) x), \tag{1.4} \end{equation*}(1.4)a(v)u=u(L(a(v))x),
where L ( a ( v ) ) x L ( a ( v ) ) x L(a(v))xL(a(v)) xL(a(v))x is the Lorentz-transformed vector and u u uuu is the normalized spinor.
Problem 1.36 Use Einstein's postulate to derive the expressions for a Lorentz boost in the x x xxx-direction.
Problem 1.37 In an inertial frame S S SSS, rockets A A AAA and B B BBB traveling with velocities v v vvv and v v -v-vv, respectively, pass each other at time t = 0 t = 0 t=0t=0t=0 at the spatial origin. A time t 0 t 0 t_(0)t_{0}t0 later, light signals are sent from the origin toward each of the spaceships. Compute the time difference between the spaceships receiving the light signals in the rest frame of one of the rockets.

1.4 Relativistic Velocities and Proper Quantities

Problem 1.38 a) Explain the concept of "relativity of simultaneity." Illustrate it in a spacetime diagram.
b) The worldline of a massive particle in Minkowski space is described by the following equations in some inertial frame ( x μ ) = ( c t , x , y , z ) x μ = ( c t , x , y , z ) (x^(mu))=(ct,x,y,z)\left(x^{\mu}\right)=(c t, x, y, z)(xμ)=(ct,x,y,z),
(1.5) x ( t ) = 3 2 a t 2 , y ( t ) = 2 a t 2 , z ( t ) = 0 (1.5) x ( t ) = 3 2 a t 2 , y ( t ) = 2 a t 2 , z ( t ) = 0 {:(1.5)x(t)=(3)/(2)at^(2)","quad y(t)=2at^(2)","quad z(t)=0:}\begin{equation*} x(t)=\frac{3}{2} a t^{2}, \quad y(t)=2 a t^{2}, \quad z(t)=0 \tag{1.5} \end{equation*}(1.5)x(t)=32at2,y(t)=2at2,z(t)=0
where a a aaa is constant and 0 t t 0 0 t t 0 0 <= t <= t_(0)0 \leq t \leq t_{0}0tt0 for some value of t 0 t 0 t_(0)t_{0}t0. Compute the particle's 4 -velocity and 4 -acceleration components. What values of t 0 t 0 t_(0)t_{0}t0 are possible and why? Compute the proper time along this worldline from t = 0 t = 0 t=0t=0t=0 to t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0.
Problem 1.39 A rod moves with velocity v v vvv along the positive x x xxx-axis in an inertial frame S S SSS. An observer at rest in S S SSS measures the length of the rod to be L L LLL. Another observer moves with the velocity v v -v-vv along the x x xxx-axis. What length, expressed as a function of L L LLL and v v vvv, will this observer measure for the rod? The measurement is done as usual with the endpoints being measured simultaneously for each observer in their respective frames.
Problem 1.40 The worldline of a particle is described by the coordinates x μ ( t ) x μ ( t ) x^(mu)(t)x^{\mu}(t)xμ(t) in the system S S SSS. An observer at rest in the system S S S^(')S^{\prime}S, with velocity u u uuu along the positive x 2 x 2 x^(2)x^{2}x2-axis relative to S S SSS, measures the velocity of the particle at time t t t^(')t^{\prime}t. Express his result as a function of the velocity of the particle in S S SSS and u u uuu.
Problem 1.41 A spaceship is moving away from Earth. The effect of the engines is regulated so that the the passengers feel the constant acceleration g g ggg. Calculate the distance between the Earth and the spaceship (measured in the rest frame of the Earth) as a function of
a) the time on Earth.
b) the time on the spaceship.
The commander of the spaceship is 40 years of age at the beginning of the voyage. How old is (s)he when the spaceship reaches the Andromeda Galaxy, which lies about 2500000 light years away from Earth?
Hint: 1 year π 10 7 s π 10 7 s ~~pi*10^(7)s\approx \pi \cdot 10^{7} \mathrm{~s}π107 s and g 10 m / s 2 g 10 m / s 2 g~~10m//s^(2)g \approx 10 \mathrm{~m} / \mathrm{s}^{2}g10 m/s2.
Problem 1.42 A rocket (with rest mass m 0 m 0 m_(0)m_{0}m0 ) starts from rest at the origin of a coordinate system K K KKK. Its velocity along the positive x x xxx-axis is increased by shooting
matter from the rocket with constant velocity w w www relative to the instantaneous rest frame of the rocket in the negative x x xxx-direction. Compute the remaining mass m m mmm of the rocket as a function of its velocity v v vvv with respect to the origin of K K KKK.
Problem 1.43 You and your friend are in intergalactic space (assume the Minkowski metric). You leave simultaneously from a space station, with equal speeds v v vvv, in orthogonal directions. Neglect acceleration. After a time T T TTT has passed in your inertial frame, you want to send a message to your friend using a light signal. In which direction (in your rest frame) should you send it?
Problem 1.44 a) In an inertial frame S S SSS, an object travels with 3-velocity u u uuu. A different inertial frame S S S^(')S^{\prime}S is moving in the negative x x xxx-direction relative to S S SSS with relative speed v v v^(')v^{\prime}v. Write down the 4 -velocities of the object and S S S^(')S^{\prime}S in S S SSS.
b) Show that the gamma factor of an object in any inertial frame is given by the inner product of the 4 -velocity of the object and the 4 -velocity of an object at rest in the frame.
c) Express the gamma factor of the object in a) in the frame S S S^(')S^{\prime}S using the result from b).
Problem 1.45 Show that the 4-velocity V μ = d x μ / d τ V μ = d x μ / d τ V^(mu)=dx^(mu)//d tauV^{\mu}=d x^{\mu} / d \tauVμ=dxμ/dτ and 4-acceleration A μ = A μ = A^(mu)=A^{\mu}=Aμ= d V μ / d τ d V μ / d τ dV^(mu)//d taud V^{\mu} / d \taudVμ/dτ of an object are always perpendicular, where τ τ tau\tauτ is the proper time of the object such that V 2 = 1 V 2 = 1 V^(2)=1V^{2}=1V2=1.
Problem 1.46 You are traveling in your spaceship in flat intergalactic space such that special relativity can be used. You are on your way to a space station when you suddenly discover an enemy spaceship on your radar. You immediately send out a light signal for help to the space station. When you send out your signal, the distance in your coordinate system to the space station is 1 light day. You have a relative speed of c / 4 c / 4 c//4c / 4c/4 toward the space station. When the space station receives your signal, they send out a rescue spaceship with a speed 3 c / 4 3 c / 4 3c//43 c / 43c/4 relative the space station. How long does it take before you get help (as measured by your own clock)? First, how long does it take your light signal to reach the space station, and second, how long does it take for the rescue spaceship to reach you?
Problem 1.47 On an interstellar highway there is a speed limit of u u uuu relative to a reference frame S S SSS. A member of the intergalactic police force is at rest in this system when a spaceship passes at constant velocity v > u v > u v > uv>uv>u (see Figure 1.5). Eager to do the job properly, the police officer starts the pursuit, accelerating with constant proper acceleration a a aaa. The pursuit ends when the police officer catches up with the criminal.
a) How long does the pursuit take according to the criminal?
b) How long does the pursuit take according to the police officer?
c) What is the relative velocity between the police officer and the criminal at the end of the pursuit?
Problem 1.48 An astronaut on an accelerated spaceship uses a coordinate system ( T , X , Y , Z T , X , Y , Z T,X,Y,ZT, X, Y, ZT,X,Y,Z ) related to an inertial system ( t , x , y , z ( t , x , y , z (t,x,y,z(t, x, y, z(t,x,y,z ) as follows (we set c = 1 c = 1 c=1c=1c=1 )
(1.6) t = X sinh ( a T ) , x = X cosh ( a T ) , y = Y , z = Z (1.6) t = X sinh ( a T ) , x = X cosh ( a T ) , y = Y , z = Z {:(1.6)t=X sinh(aT)","quad x=X cosh(aT)","quad y=Y","quad z=Z:}\begin{equation*} t=X \sinh (a T), \quad x=X \cosh (a T), \quad y=Y, \quad z=Z \tag{1.6} \end{equation*}(1.6)t=Xsinh(aT),x=Xcosh(aT),y=Y,z=Z
Figure 1.5 A space ship passing the intergalactic police at a relative constant velocity v v vvv.
a) Compute the metric tensor in the astronaut's coordinate system. (The metric in the inertial system is, of course, d s 2 = d t 2 d x 2 d y 2 d z 2 d s 2 = d t 2 d x 2 d y 2 d z 2 ds^(2)=dt^(2)-dx^(2)-dy^(2)-dz^(2)d s^{2}=d t^{2}-d x^{2}-d y^{2}-d z^{2}ds2=dt2dx2dy2dz2.)
b) Let ( k μ ) = ( ω , ω cos ( θ ) , 0 , ω sin ( θ ) ) k μ = ( ω , ω cos ( θ ) , 0 , ω sin ( θ ) ) (k^(mu))=(omega,omega cos(theta),0,omega sin(theta))\left(k^{\mu}\right)=(\omega, \omega \cos (\theta), 0, \omega \sin (\theta))(kμ)=(ω,ωcos(θ),0,ωsin(θ)) be the 4 -wavevector of a photon emitted at time t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0 at the position x = x 0 , y = z = 0 x = x 0 , y = z = 0 x=x_(0),y=z=0x=x_{0}, y=z=0x=x0,y=z=0 in the abovementioned inertial system. Compute the components of this 4 -wavevector in the astronaut's coordinate system.
c) Compute the duration of a trip of the spaceship on the astronaut's watch (i.e., the proper time) if the trajectory of his spaceship on this trip is, in the astronaut's coordinate system, X ( T ) = X 0 = X ( T ) = X 0 = X(T)=X_(0)=X(T)=X_{0}=X(T)=X0= constant, Y ( T ) = v T Y ( T ) = v T Y(T)=vTY(T)=v TY(T)=vT for some constant v > 0 v > 0 v > 0v>0v>0, Z ( T ) = 0 Z ( T ) = 0 Z(T)=0Z(T)=0Z(T)=0, and 0 T T 0 0 T T 0 0 <= T <= T_(0)0 \leq T \leq T_{0}0TT0.
Problem 1.49 A rocket A A AAA is accelerating with constant proper acceleration α α alpha\alphaα such that its worldline is given by t 2 x 2 = 1 / α 2 t 2 x 2 = 1 / α 2 t^(2)-x^(2)=-1//alpha^(2)t^{2}-x^{2}=-1 / \alpha^{2}t2x2=1/α2 in the reference frame S S SSS. Assume that S S S^(')S^{\prime}S is a different reference frame related to S S SSS by a Lorentz transformation in standard configuration with velocity v v vvv. What is the coordinate acceleration a a a^(')a^{\prime}a in the system S S S^(')S^{\prime}S at time t = 0 t = 0 t^(')=0t^{\prime}=0t=0 ?
Problem 1.50 Two observers A A AAA and B B BBB are initially colocated at rest in the inertial system S S SSS. At a given time in S S SSS, observer B B BBB starts accelerating with a proper acceleration α α alpha\alphaα. A time t 0 t 0 t_(0)t_{0}t0 later (as measured by A A AAA ), a light signal is sent from A A AAA toward B B BBB. Find an expression for the proper timed elapsed for observer B B BBB when B B BBB receives the signal. Discuss the limiting cases.
Problem 1.51 Particles in a circular accelerator are accelerated by an electromagnetic field in such a way that they are kept in a circular orbit with constant velocity. What are the corresponding 4-acceleration and proper acceleration of the particle and what is the eigentime required for the particles to complete one orbit? Introduce any quantities required to solve the problem.
Problem 1.52 An object of internal energy M M MMM moving with 4-velocity V V VVV is being acted upon by a force F = f U F = f U F=fUF=f UF=fU, where the known 4-vector U U UUU fulfills U 2 = 1 U 2 = 1 U^(2)=1U^{2}=1U2=1 and f f fff is a scalar. How fast is the internal energy of the object increasing (with respect to its proper time) and what is the proper acceleration of the object?
Problem 1.53 Consider a 4-force F μ = ( 0 , f ) μ F μ = ( 0 , f ) μ F^(mu)=(0,f)^(mu)F^{\mu}=(0, \boldsymbol{f})^{\mu}Fμ=(0,f)μ acting on an object of rest energy m m mmm with 3-velocity v v v\boldsymbol{v}v. Compute the rate of change in the rest energy d m / d τ d m / d τ dm//d taud m / d \taudm/dτ and the
Figure 1.6 An observer o o ooo moving at velocity v v vvv and a particle p p ppp moving toward the observer o o ooo at velocity u u uuu in a direction that makes an angle θ θ theta\thetaθ with the direction of the velocity v v vvv.
proper acceleration α α alpha\alphaα, where τ τ tau\tauτ is the proper time of the object worldline. Discuss the requirements on f f fff for the 4 -force to be a pure force.
Problem 1.54 Given an object with acceleration a 0 a 0 a_(0)\boldsymbol{a}_{0}a0 in its instantaneous rest frame S S SSS, find an expression for the acceleration in the inertial frame S S S^(')S^{\prime}S, which is moving in the x x xxx-direction with velocity v v vvv relative to S S SSS. What is the maximal and minimal acceleration in S S S^(')S^{\prime}S depending on the direction of the acceleration based on your expression?
Problem 1.55 A particle has 4-momentum P = ( E , p ) P = ( E , p ) P=(E,p)P=(E, p)P=(E,p) in an inertial frame S S SSS. An observer is moving by with velocity v v vvv in the x x xxx-direction. Compute the total energy this observer will measure for the particle and the velocity of the particle in the x x x^(')x^{\prime}x-direction of the observer's rest frame.
Problem 1.56 An object originally at rest with mass m ( 0 ) = m 0 m ( 0 ) = m 0 m(0)=m_(0)m(0)=m_{0}m(0)=m0 is affected by a constant 4-force F μ = f ( 1 , 1 ) μ F μ = f ( 1 , 1 ) μ F^(mu)=f(1,1)^(mu)F^{\mu}=f(1, \mathbf{1})^{\mu}Fμ=f(1,1)μ. Find the object's mass and the time elapsed in the initial rest frame as a function of the object's proper time τ τ tau\tauτ.
Problem 1.57 An observer o o ooo is moving at velocity v v vvv in the x x xxx-direction in an inertial frame S S SSS. In the same inertial frame, a particle p p ppp hits the observer while traveling at a speed u u uuu in a direction that makes an angle θ θ theta\thetaθ with the negative x x xxx direction, see Figure 1.6. What is the speed and angle that the observer will measure for the particle? Verify that your result is consistent with both the ultrarelativistic and nonrelativistic limits, i.e., u 1 u 1 u rarr1u \rightarrow 1u1 and u , v 1 u , v 1 u,v≪1u, v \ll 1u,v1, respectively.
Problem 1.58 In an inertial frame S S SSS an object starting at rest at t = 0 t = 0 t=0t=0t=0 is moving with constant coordinate acceleration a a aaa, i.e., x = a t 2 / 2 x = a t 2 / 2 x=at^(2)//2x=a t^{2} / 2x=at2/2. Determine the proper time for the object to reach the speed v 0 v 0 v_(0)v_{0}v0 in S S SSS and the proper acceleration of the object as a function of the time t t ttt in S S SSS.

1.5 Relativistic Optics

Problem 1.59 In 1851, Fizeau measured the speed of light in running water. His result can be summarized in the formula
(1.7) u = u 0 + k v (1.7) u = u 0 + k v {:(1.7)u=u_(0)+kv:}\begin{equation*} u=u_{0}+k v \tag{1.7} \end{equation*}(1.7)u=u0+kv
where u u uuu is the speed of light in water, that runs with velocity v v vvv. The speed of light in water at rest is u 0 u 0 u_(0)u_{0}u0 and the drag coefficient k k kkk is given by
(1.8) k = 1 1 n 2 (1.8) k = 1 1 n 2 {:(1.8)k=1-(1)/(n^(2)):}\begin{equation*} k=1-\frac{1}{n^{2}} \tag{1.8} \end{equation*}(1.8)k=11n2
where n = c / u 0 n = c / u 0 n=c//u_(0)n=c / u_{0}n=c/u0 is the refractive index of water. Explain Fizeau's result!
Problem 1.60 See Problem 1.59. Is Fizeau's result still valid if the water runs perpendicular to the motion of light? If not, what is the correction?
Problem 1.61 In 1965, Maarten Schmidt at the Mount Palomar Observatory could identify the strongly redshift Lyman α α alpha\alphaα line in the spectrum of the quasi-stellar radio source 3C 9. Normally, this line has the wavelength 1215 1215 1215"Å"1215 \AA1215. Schmidt instead found the value 3600 3600 3600"Å"3600 \AA3600 for this line in this radio source. It is possible to explain the redshift in terms of the Doppler effect. This would imply that 3C 9 moves with an enormous speed relative to our galaxy. Determine a lower bound for the speed of 3C 9 .
Problem 1.62 A plane electromagnetic wave moving along the x 1 x 1 x^(1)x^{1}x1-axis has the form
(1.9) E ( x ) = E 0 sin [ 2 π ( x 1 λ v t ) ] (1.9) E ( x ) = E 0 sin 2 π x 1 λ v t {:(1.9)E(x)=E_(0)sin[2pi((x^(1))/(lambda)-vt)]:}\begin{equation*} E(x)=E_{0} \sin \left[2 \pi\left(\frac{x^{1}}{\lambda}-v t\right)\right] \tag{1.9} \end{equation*}(1.9)E(x)=E0sin[2π(x1λvt)]
Introduce the angular frequency ω = 2 π ν ω = 2 π ν omega=2pi nu\omega=2 \pi \nuω=2πν and show that the argument of the wave can be written in the form x μ k μ x μ k μ -x_(mu)k^(mu)-x_{\mu} k^{\mu}xμkμ, where k = ( ω c , ω c , 0 , 0 ) k = ω c , ω c , 0 , 0 k=((omega )/(c),(omega )/(c),0,0)k=\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right)k=(ωc,ωc,0,0) is the 4 -wave vector of the light wave traveling along the positive x 1 x 1 x^(1)x^{1}x1-axis. Show that this vector is lightlike and deduce the formula for the Doppler shift by calculating the change in angular frequency ω ω omega\omegaω under a Lorentz transformation along the x 1 x 1 x^(1)x^{1}x1-axis. What does the formula for the Doppler shift look like expressed in terms of the rapidity θ θ theta\thetaθ ?
Problem 1.63 A gamma ray burst (GRB) observed in a cluster of faraway galaxies is time dilated and therefore has a total duration about twice as long as GRBs in nearby galaxies. According to the Hubble law, the recession speed is proportional to the distance to the GRB. Calculate the Doppler redshift z = Δ λ / λ 0 z = Δ λ / λ 0 z=Delta lambda//lambda_(0)z=\Delta \lambda / \lambda_{0}z=Δλ/λ0 of a typical spectral line from the distant GRB, where λ λ lambda\lambdaλ and λ 0 λ 0 lambda_(0)\lambda_{0}λ0 are the observed and emitted wavelengths, respectively.
Hint: All GRBs can be considered to have the same duration when measured in their respective rest frames.
Problem 1.64 A person watches two objects with constant velocities on a collision course, i.e., they approach each other on a straight line.
a) Assuming that both objects' velocities have the same absolute value c / 2 c / 2 c//2c / 2c/2 in the person's frame of reference, compute the absolute value of the velocity with which a person traveling with the first object sees the other object approaching.
b) Assume that the first object sends a light pulse from a ruby laser, which produces visible light with a wavelength λ 0 = 694.3 nm λ 0 = 694.3 nm lambda_(0)=694.3nm\lambda_{0}=694.3 \mathrm{~nm}λ0=694.3 nm, toward the second object. Compute the wavelength of this light pulse as seen by an observer on the second object.
Problem 1.65 A light source is moving at speed v v vvv and at angle θ θ theta\thetaθ relative to the separation between the source and a stationary observer.
a) Consider a light pulse with frequency ω 0 ω 0 omega_(0)\omega_{0}ω0 in the rest frame of the source and determine the frequency ω ω omega\omegaω measured by the observer.
b) Compute the angle θ θ theta\thetaθ for which ω = ω 0 ω = ω 0 omega=omega_(0)\omega=\omega_{0}ω=ω0.
Problem 1.66 A large disk rotates at uniform angular speed Ω Ω Omega\OmegaΩ in an inertial frame S S SSS. Two observers, O 1 O 1 O_(1)O_{1}O1 and O 2 O 2 O_(2)O_{2}O2, ride on the disk at radial distances r 1 r 1 r_(1)r_{1}r1 and r 2 r 2 r_(2)r_{2}r2, respectively, from the center (not necessarily on the same radial line). They carry clocks, C 1 C 1 C_(1)C_{1}C1 and C 2 C 2 C_(2)C_{2}C2, which they adjust so that the clocks keep time with clocks in S S SSS, i.e., the clocks speed up their natural rates by the Lorentz factors
(1.10) γ 1 = 1 1 r 1 2 Ω 2 / c 2 ; γ 2 = 1 1 r 2 2 Ω 2 / c 2 , (1.10) γ 1 = 1 1 r 1 2 Ω 2 / c 2 ; γ 2 = 1 1 r 2 2 Ω 2 / c 2 , {:(1.10)gamma_(1)=(1)/(sqrt(1-r_(1)^(2)Omega^(2)//c^(2)));quadgamma_(2)=(1)/(sqrt(1-r_(2)^(2)Omega^(2)//c^(2)))",":}\begin{equation*} \gamma_{1}=\frac{1}{\sqrt{1-r_{1}^{2} \Omega^{2} / c^{2}}} ; \quad \gamma_{2}=\frac{1}{\sqrt{1-r_{2}^{2} \Omega^{2} / c^{2}}}, \tag{1.10} \end{equation*}(1.10)γ1=11r12Ω2/c2;γ2=11r22Ω2/c2,
respectively. By the stationary nature of the situation, C 2 C 2 C_(2)C_{2}C2 cannot appear to gain or lose relative to C 1 C 1 C_(1)C_{1}C1. Deduce that, when O 2 O 2 O_(2)O_{2}O2 sends a light signal to O 1 O 1 O_(1)O_{1}O1, this signal is affected by a Doppler shift ω 2 / ω 1 = γ 2 / γ 1 ω 2 / ω 1 = γ 2 / γ 1 omega_(2)//omega_(1)=gamma_(2)//gamma_(1)\omega_{2} / \omega_{1}=\gamma_{2} / \gamma_{1}ω2/ω1=γ2/γ1.
Note that, in particular, there is no relative Doppler shift between any two observers equidistant from the center.
Problem 1.67 A light source is moving with speed v v vvv through an optical medium with refractive index n n nnn. Derive an expression for the ratio between the frequency in the frame of the medium and the frequency in the frame of the source as a function of v , n v , n v,nv, nv,n, and the angle θ θ theta\thetaθ between the movement direction of the source and the propagation direction of the light (in the frame of the medium).
Problem 1.68 In an inertial frame S S SSS, a mirror is oriented perpendicular to the x x xxx axis and moving with velocity v v vvv in the x x xxx-direction, see Figure 1.7. A light pulse with frequency ω ω omega\omegaω approaches the mirror at an angle θ in θ in  theta_("in ")\theta_{\text {in }}θin  in S S SSS. What is the scattering angle θ out θ out  theta_("out ")\theta_{\text {out }}θout  and what frequency does the outgoing light have? Explain what happens when v < cos θ in v < cos θ in  v < -cos theta_("in ")v<-\cos \theta_{\text {in }}v<cosθin  ?
Problem 1.69 In an inertial frame S S SSS, a light pulse is being directed at an optical medium with refractive index n n nnn which is moving with velocity v v vvv orthogonal to its surface, see Figure 1.8. In the rest frame of the medium, the light pulse is refracted according to Snell's law. An observer in S S SSS makes the interesting observation that the light pulse is still traveling in the same direction after entering the medium. Compute the index of refraction for the medium in terms of the velocity v v vvv and the angle θ θ theta^(')\theta^{\prime}θ between the initial direction of the light pulse and the direction of motion for S S SSS in the rest frame of the medium.
Figure 1.7 A mirror perpendicular to the x x xxx-axis moving with velocity v v vvv in the x x xxx-direction. The ingoing light has frequency ω ω omega\omegaω and makes an angle θ in θ in  theta_("in ")\theta_{\text {in }}θin  with the x x xxx-axis, whereas the outgoing light has frequency ω ω omega^(')\omega^{\prime}ω and makes an angle θ out θ out  theta_("out ")\theta_{\text {out }}θout  with the x x xxx-axis.
Figure 1.8 An optical medium with refractive index n n nnn moving with velocity v v vvv orthogonal to its surface. In the inertial frame S S SSS, an observer makes the observation that a light pulse is traveling in the same direction (at an angle θ θ theta\thetaθ relative to the velocity v v vvv ) before and after entering the medium.
Problem 1.70 A sine wave propagating in a medium can be described by the function sin ( N x ) sin ( N x ) sin(N*x)\sin (N \cdot x)sin(Nx) in the medium rest frame. Here, ( N μ ) = ( ω , k ) N μ = ( ω , k ) (N^(mu))=(omega,k)\left(N^{\mu}\right)=(\omega, k)(Nμ)=(ω,k), where ω ω omega\omegaω is the angular frequency and k k kkk the wave number. Assuming the wave velocity in the medium is u u uuu, the relationship between k k kkk and ω ω omega\omegaω is k u = ω k u = ω ku=omegak u=\omegaku=ω. A source with internal frequency ω 0 ω 0 omega_(0)\omega_{0}ω0 is moving through the medium with velocity v v vvv. Compute the Doppler shifted frequency ω ω omega\omegaω in the medium rest frame when the waves are traveling in the direction of motion. Also discuss the special cases v = u v = u v=uv=uv=u and v = u v = u v=-uv=-uv=u and make sure that your solution reduces to the classical Doppler formula when v 1 v 1 v≪1v \ll 1v1.
Problem 1.71 Using the same setup as in Problem 1.50 and assuming that A A AAA sends the light signal using a frequency ω ω omega\omegaω, compute the frequency observed by B B BBB. In addition, if B B BBB carries a mirror and reflects the signal back at A A AAA, find the frequency observed by A A AAA for the reflected signal.

1.6 Relativistic Mechanics

Problem 1.72 The rest energy of an electron is about 0.51 MeV , i.e., the energy a charged particle, with charge equal to the electron charge, would receive when falling down a potential difference of 0.51 MV . Assuming that the electron is accelerated through a linear accelerator (starting from rest) with a potential difference of 10 6 V 10 6 V 10^(6)V10^{6} \mathrm{~V}106 V. Compute the final velocity of the electron.
Problem 1.73 a) An electron e e e^(-)e^{-}e(with mass m e m e m_(e)m_{e}me ) collides with a positron e + e + e^(+)e^{+}e+(i.e., the antiparticle of the electron with the same mass m e m e m_(e)m_{e}me as the electron). Show that they cannot annihilate into a single photon γ γ gamma\gammaγ (a photon has zero mass), i.e., the process e + e + γ e + e + γ e^(-)+e^(+)longrightarrow gammae^{-}+e^{+} \longrightarrow \gammae+e+γ is impossible due to conservation of energy and momentum.
b) Also show that an electron cannot spontaneously emit a photon.
c) Can an electron colliding with a positron annihilate into two photons? Justify your answer.
Problem 1.74 An elementary particle with mass M M MMM decays into two particles a a aaa and b b bbb with masses m a m a m_(a)m_{a}ma and m b m b m_(b)m_{b}mb, respectively. Calculate the momentum of particle a a aaa in the rest frame of particle b b bbb.
Problem 1.75 A particle A A AAA with mass m A m A m_(A)m_{A}mA decays into two particles B B BBB and C C CCC with masses m B m B m_(B)m_{B}mB and m C m C m_(C)m_{C}mC, respectively. Assume that particle A A AAA has speed v A v A v_(A)v_{A}vA before the decay and that particle B B BBB is at rest after the decay, i.e., p B = 0 p B = 0 p_(B)=0\boldsymbol{p}_{B}=\mathbf{0}pB=0. Express the speed v A v A v_(A)v_{A}vA in the masses m A , m B m A , m B m_(A),m_(B)m_{A}, m_{B}mA,mB, and m C m C m_(C)m_{C}mC.
Problem 1.76 Two particles, 1 and 2, with masses m 1 m 1 m_(1)m_{1}m1 and m 2 m 2 m_(2)m_{2}m2, respectively, collide and form a new particle with mass M M MMM. Calculate the mass M M MMM and the velocity v v v\boldsymbol{v}v of this new particle in the rest frame of particle 2 as a function of the velocity v 1 v 1 v_(1)\boldsymbol{v}_{1}v1 of particle 1 in the rest frame of particle 2 and the masses m 1 m 1 m_(1)m_{1}m1 and m 2 m 2 m_(2)m_{2}m2.
Problem 1.77 a) Two particles with rest masses m 1 m 1 m_(1)m_{1}m1 and m 2 m 2 m_(2)m_{2}m2, respectively, move along the x x xxx-axis in the inertial frame of some observer at uniform velocities u 1 u 1 u_(1)u_{1}u1 and u 2 u 2 u_(2)u_{2}u2, respectively. They collide and form a single particle with rest mass m m mmm moving at uniform velocity u u uuu. Assuming that c = 1 c = 1 c=1c=1c=1, prove that
(1.11) m 2 = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( u 1 ) γ ( u 2 ) ( 1 u 1 u 2 ) , (1.11) m 2 = m 1 2 + m 2 2 + 2 m 1 m 2 γ u 1 γ u 2 1 u 1 u 2 , {:(1.11)m^(2)=m_(1)^(2)+m_(2)^(2)+2m_(1)m_(2)gamma(u_(1))gamma(u_(2))(1-u_(1)u_(2))",":}\begin{equation*} m^{2}=m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma\left(u_{1}\right) \gamma\left(u_{2}\right)\left(1-u_{1} u_{2}\right), \tag{1.11} \end{equation*}(1.11)m2=m12+m22+2m1m2γ(u1)γ(u2)(1u1u2),
and also find u u uuu.
b) Show that the above expression can be written as
(1.12) m 2 = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( v ) , (1.12) m 2 = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( v ) , {:(1.12)m^(2)=m_(1)^(2)+m_(2)^(2)+2m_(1)m_(2)gamma(v)",":}\begin{equation*} m^{2}=m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma(v), \tag{1.12} \end{equation*}(1.12)m2=m12+m22+2m1m2γ(v),
when v v vvv is the velocity of particle 2 as measured in the rest frame of particle 1 .
c) Consider two different situations and in both of the situations the relative velocity v v vvv as defined above is the same, and thus, the rest mass m m mmm is the same in both situations, but in one u 1 = 0 u 1 = 0 u_(1)=0u_{1}=0u1=0 and in the other m 1 γ ( u 1 ) u 1 = m 2 γ ( u 2 ) u 2 m 1 γ u 1 u 1 = m 2 γ u 2 u 2 m_(1)gamma(u_(1))u_(1)=-m_(2)gamma(u_(2))u_(2)m_{1} \gamma\left(u_{1}\right) u_{1}=-m_{2} \gamma\left(u_{2}\right) u_{2}m1γ(u1)u1=m2γ(u2)u2. What is the difference in total energy for the two situations in the frame of the observer?
Problem 1.78 A pion with mass m π m π m_(pi)m_{\pi}mπ and energy E π E π E_(pi)E_{\pi}Eπ moves along the x x xxx-axis. It decays into a muon with mass m μ m μ m_(mu)m_{\mu}mμ and a neutrino with approximately zero mass. Calculate the energy E μ E μ E_(mu)E_{\mu}Eμ of the muon when it moves at a right angle relative to the x x xxx-axis in terms of the velocity of the incoming pion and the masses.
Problem 1.79 A pion with mass m π m π m_(pi)m_{\pi}mπ decays into an electron with mass m m mmm and an antineutrino with mass m ν m ν m_(nu)m_{\nu}mν. Calculate the velocity of the antineutrino in the rest frame of the electron as a function of the masses of the particles and determine the limiting value of this velocity as the mass of the antineutrino goes to zero.
Problem 1.80 In June 1998, the Super-Kamiokande Collaboration in Japan reported that it had found evidence for massive neutrinos. Super-Kamiokande measures so-called atmospheric neutrinos, which are produced in hadronic showers resulting from collisions of cosmic rays with nuclei in the upper atmosphere. Two of the dominating processes in the production of atmospheric neutrinos are
π + μ + + v μ , π + μ + + v μ , pi^(+)longrightarrowmu^(+)+v_(mu),\pi^{+} \longrightarrow \mu^{+}+v_{\mu},π+μ++vμ,
where π + π + pi^(+)\pi^{+}π+is a pion, μ + μ + mu^(+)\mu^{+}μ+is an antimuon, and v μ v μ v_(mu)v_{\mu}vμ is a muon neutrino, followed by
μ + e + + v ¯ μ + v e μ + e + + v ¯ μ + v e mu^(+)longrightarrowe^(+)+ bar(v)_(mu)+v_(e)\mu^{+} \longrightarrow e^{+}+\bar{v}_{\mu}+v_{e}μ+e++v¯μ+ve
where e + e + e^(+)e^{+}e+is a positron, v ¯ μ v ¯ μ bar(v)_(mu)\bar{v}_{\mu}v¯μ is an antimuon neutrino, and v e v e v_(e)v_{e}ve is an electron neutrino.
a) Calculate the kinetic energy of the antimuon, T μ + T μ + T_(mu^(+))T_{\mu^{+}}Tμ+, and the absolute value of the 3 -momentum of the muon neutrino, p ν μ p ν μ p_(nu_(mu))p_{\nu_{\mu}}pνμ, when the pion decays at rest according to the first decay. Despite the small mass of the muon neutrino, neglect it! The mass of the pion is m π m π m_(pi)m_{\pi}mπ and the mass of the antimuon is m μ m μ m_(mu)m_{\mu}mμ.
b) How far will one of the antimuons, which are produced in the first decay, travel (on average) in the pion rest frame before it decays according to the second decay? The mean lifetime of an antimuon at rest is τ μ τ μ tau_(mu)\tau_{\mu}τμ.
Problem 1.81 The pions in the sky that are decaying into muons as in Problem 1.14 are produced in collisions between protons in the primary cosmic rays and nitrogen or oxygen in the air. When a pion with energy of 2 GeV is produced, what energy does the muon have if it continues in the same direction as the pion? The expression can be simplified due to the high energy of the pion. What is the resulting expression? What is the muon energy? The pion has a rest mass of 140 MeV , and the neutrino mass can be neglected.
Problem 1.82 A beam of protons that are accelerated to a very high energy hits a beryllium target and produces a shower of particles. Two detectors are placed in a plane behind the target symmetrically around the proton beam axis. Each detector makes an angle of 45 45 45^(@)45^{\circ}45 with this axis and detects μ + μ μ + μ mu^(+)mu^(-)\mu^{+} \mu^{-}μ+μ-pairs, one type of particle in each detector. When the momentum of each muon is 2.2 GeV , one sees an enhancement in the muon rate. This is interpreted as the production of a resonance R R RRR of mass M R M R M_(R)M_{R}MR that decays into the muons. What is the mass M R M R M_(R)M_{R}MR of this resonance? The muon mass is 106 MeV .
Problem 1.83 A particle with mass M M MMM and 4-momentum p = ( E , p ) p = ( E , p ) p=(E,p)p=(E, \mathbf{p})p=(E,p) moves toward a detector when it suddenly decays and emits a photon in the direction of motion.
Figure 1.9 Scattering of two photons γ + γ γ + γ γ + γ γ + γ gamma+gamma longrightarrow gamma+gamma\gamma+\gamma \longrightarrow \gamma+\gammaγ+γγ+γ.
The energy registered by the detector is ω ω omega\omegaω. Determine what energy the photon had in the rest frame of the decaying particle.
Problem 1.84 An electron moves with constant velocity toward a positron at rest and they annihilate into two photons. The photons go out with angles ϕ ϕ phi\phiϕ and ϕ ϕ -phi-\phiϕ relative to the direction of the incoming electron.
a) Calculate the angle as a function of the total energy of the electron.
b) Show that in the nonrelativistic limit the angle is given by cos ϕ = v / ( 2 c ) cos ϕ = v / ( 2 c ) cos phi=v//(2c)\cos \phi=v /(2 c)cosϕ=v/(2c).
Problem 1.85 Two photons with wavelengths λ 1 λ 1 lambda_(1)\lambda_{1}λ1 and λ 2 λ 2 lambda_(2)\lambda_{2}λ2, respectively, are scattered against each other according to Figure 1.9. Calculate the wavelength of the photon with scattering angle θ θ theta\thetaθ, i.e., express λ λ lambda\lambdaλ as a function of λ 1 , λ 2 λ 1 , λ 2 lambda_(1),lambda_(2)\lambda_{1}, \lambda_{2}λ1,λ2, and θ θ theta\thetaθ.
Hint: p = h λ p = h λ p=(h)/( lambda)p=\frac{h}{\lambda}p=hλ, where h h hhh is Planck's constant.
Problem 1.86 A K-meson with mass M M MMM decays at rest into two charged pions with the same mass m m mmm and a photon according to the reaction formula
K 0 ( P ) π + ( p 1 ) + π ( p 2 ) + γ ( k ) K 0 ( P ) π + p 1 + π p 2 + γ ( k ) K^(0)(P)longrightarrowpi^(+)(p_(1))+pi^(-)(p_(2))+gamma(k)K^{0}(P) \longrightarrow \pi^{+}\left(p_{1}\right)+\pi^{-}\left(p_{2}\right)+\gamma(k)K0(P)π+(p1)+π(p2)+γ(k)
The momenta of the particles are given in parentheses after each particle symbol. Calculate the speed v v vvv of the pions in center-of-mass frame where (where p 1 + p 2 = 0 p 1 + p 2 = 0 p_(1)+p_(2)=0\mathbf{p}_{1}+\mathbf{p}_{2}=0p1+p2=0 ) as a function of the masses of the particles and the photon energy k 0 = ω k 0 = ω k^(0)=omegak^{0}=\omegak0=ω in the rest frame of the decaying particle.
Problem 1.87 A Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 particle with speed c / 3 c / 3 c//3c / 3c/3 in the direction toward a gamma detector suddenly decays into a Λ Λ Lambda\LambdaΛ particle and a photon. The photon continues toward the detector.
a) What energy does the Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 particle have in the system in which the detector is at rest?
b) What energy does the photon have in the rest system of the Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 particle?
c) What energy will be registered in the detector?
The mass of the Λ Λ Lambda\LambdaΛ is m Λ 1115.7 MeV m Λ 1115.7 MeV m_(Lambda)~~1115.7MeVm_{\Lambda} \approx 1115.7 \mathrm{MeV}mΛ1115.7MeV and that of Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 is m Σ 0 1192.6 MeV m Σ 0 1192.6 MeV m_(Sigma^(0))~~1192.6MeVm_{\Sigma^{0}} \approx 1192.6 \mathrm{MeV}mΣ01192.6MeV.
Problem 1.88 In elastic scattering of two particles onto each other, the same type of particles are present before and after the collision. Thus, in e + p e + p e + p e + p e+p longrightarrow e+pe+p \longrightarrow e+pe+pe+p elastic
scattering of electrons on protons with corresponding 4-momenta p e , p p , p e p e , p p , p e p_(e),p_(p),p_(e)^(')p_{e}, p_{p}, p_{e}^{\prime}pe,pp,pe, and p p p p p_(p)^(')p_{p}^{\prime}pp, one can form an invariant called t t ttt, defined as t = ( p e p e ) 2 t = p e p e 2 t=(p_(e)-p_(e)^('))^(2)t=\left(p_{e}-p_{e}^{\prime}\right)^{2}t=(pepe)2.
a) Show that, in the center-of-mass system defined by the total 3 -momentum being 0 0 0\mathbf{0}0, the quantity t t -t-tt equals the square of the change of the 3 -momentum, i.e., t = ( p e p e ) 2 t = p e p e 2 -t=(p_(e)-p_(e)^('))^(2)-t=\left(\mathbf{p}_{e}-\mathbf{p}_{e}^{\prime}\right)^{2}t=(pepe)2 and express this quantity in terms of the scattering angle θ θ theta\thetaθ between the incoming and outgoing electrons and the modulus of the 3-momentum | p e | p e |p_(e)|\left|\mathbf{p}_{e}\right||pe| of the incoming electron.
b) Calculate the kinetic energy, T p T p T_(p)^(')T_{p}^{\prime}Tp, of the outgoing proton in the laboratory system, where the incoming proton is at rest before the collision, in terms of the variable t t ttt.
Problem 1.89 What is the kinetic energy T T TTT of the pion required to create the resonance Δ ( 1232 ) Δ ( 1232 ) Delta(1232)\Delta(1232)Δ(1232) in the reaction
π + p π + Δ π + p π + Δ pi+p longrightarrow pi+Delta\pi+p \longrightarrow \pi+\Deltaπ+pπ+Δ
where π π pi\piπ is a pion and p p ppp is a proton? The proton is at rest before the collision. The result should be expressed in terms of the masses of the particles involved.
Problem 1.90 The scattering probabilities for the reactions π + d p + p π + d p + p pi+d longrightarrow p+p\pi+d \longrightarrow p+pπ+dp+p and for the reversed reaction p + p π + d p + p π + d p+p longrightarrow pi+dp+p \longrightarrow \pi+dp+pπ+d are related due to so-called time reversal invariance. However, they must be compared at the same center-of-mass energy. Calculate the relation between the kinetic energy T π T π T_(pi)T_{\pi}Tπ of the pion ( π ) ( π ) (pi)(\pi)(π), in the frame where the deuteron ( d ) ( d ) (d)(d)(d) is at rest before the collision in the first reaction, and the kinetic energy T p T p T_(p)T_{p}Tp of one of the protons ( p p ppp ) in reversed reaction, when the other proton is at rest, respecting the above condition on the center-of-mass energy.
Problem 1.91 Consider the reaction π + + n K + + Λ π + + n K + + Λ pi^(+)+n longrightarrowK^(+)+Lambda\pi^{+}+n \longrightarrow K^{+}+\Lambdaπ++nK++Λ in the rest frame of n n nnn. The masses of the particles are m π + , m n , m K + m π + , m n , m K + m_(pi^(+)),m_(n),m_(K^(+))m_{\pi^{+}}, m_{n}, m_{K^{+}}mπ+,mn,mK+, and m Λ m Λ m_(Lambda)m_{\Lambda}mΛ, respectively. What is the kinetic energy T T TTT of the π + π + pi^(+)\pi^{+}π+when the K + K + K^(+)K^{+}K+has total energy E E EEE and moves off at an angle of 90 90 90^(@)90^{\circ}90 to the direction of the incident π + π + pi^(+)\pi^{+}π+? ( T T (T:}\left(T\right.(T should be expressed in m π + , m n m π + , m n m_(pi^(+)),m_(n)m_{\pi^{+}}, m_{n}mπ+,mn, m K + , m Λ m K + , m Λ m_(K^(+)),m_(Lambda)m_{K^{+}}, m_{\Lambda}mK+,mΛ, and E E EEE.)
Problem 1.92 The mass of the meson π 0 π 0 pi^(0)\pi^{0}π0 can be measured by the reaction
p + π π 0 + n p + π π 0 + n p+pi^(-)longrightarrowpi^(0)+np+\pi^{-} \longrightarrow \pi^{0}+np+ππ0+n
where p p ppp is a proton, π π pi^(-)\pi^{-}πis a negative pion, and n n nnn is a neutron. The uncharged π 0 π 0 pi^(0)\pi^{0}π0 meson decays very quickly into two photons and cannot be easily measured. However, the velocity of the final neutron can be measured and is found to be v n = v n = v_(n)=v_{n}=vn= ( 0.89418 ± 0.00017 ) cm / ns 0.89418 ± 0.00017 ) cm / ns 0.89418+-0.00017)cm//ns0.89418 \pm 0.00017) \mathrm{cm} / \mathrm{ns}0.89418±0.00017)cm/ns. Derive the formula that expresses the mass of the π 0 π 0 pi^(0)\pi^{0}π0 meson as a function of the masses of the proton, the π π pi^(-)\pi^{-}π, the neutron, and the velocity v n v n v_(n)v_{n}vn, assuming that the reaction takes place at rest for the incoming particles. Simplify the result by showing that the velocity is small, so that we need to retain only lowest nontrivial order in v n / c v n / c v_(n)//cv_{n} / cvn/c.
Problem 1.93 A thermal neutron is absorbed by a proton at rest and a deuteron is formed together with a photon. This exothermic reaction is formally
p + n d + γ p + n d + γ p+n longrightarrow d+gammap+n \longrightarrow d+\gammap+nd+γ
The binding energy B B BBB of the deuteron is about 2.23 MeV . Calculate, relativistically, the energy of the emitted photon as a function of the masses of the particles and the binding energy B B BBB.
Problem 1.94 A hydrogen atom H , consisting of an electron and a proton with binding energy B = 13.6 eV B = 13.6 eV B=13.6eVB=13.6 \mathrm{eV}B=13.6eV, can disintegrate into its two constituent particles by being hit by a photon. The reaction is
γ + H p + e γ + H p + e gamma+Hlongrightarrow p+e\gamma+\mathrm{H} \longrightarrow p+eγ+Hp+e
Calculate, relativistically, the smallest photon energy in the rest frame of H required for this process to occur expressed in terms of B B BBB and the hydrogen mass m H m H m_(H)m_{\mathrm{H}}mH.
Problem 1.95 Similarly to the cosmic microwave background (CMB) of photons with a temperature of T CMB 2.7 K T CMB 2.7 K T_(CMB)∼2.7KT_{\mathrm{CMB}} \sim 2.7 \mathrm{~K}TCMB2.7 K, there should be a cosmic neutrino background (CNB) with a temperature of T CNB 1.9 K T CNB 1.9 K T_(CNB)∼1.9KT_{\mathrm{CNB}} \sim 1.9 \mathrm{~K}TCNB1.9 K. At these temperatures, their kinetic energy is very tiny. Suppose a very high-energy antineutrino would hit such a neutrino and annihilate it. A result of this collision could be the production of a Z 0 Z 0 Z^(0)Z^{0}Z0 boson which decays hadronically. The reaction is formally
v ¯ + v Z 0 v ¯ + v Z 0 bar(v)+v longrightarrowZ^(0)\bar{v}+v \longrightarrow Z^{0}v¯+vZ0
What is the threshold energy for the antineutrino for this to occur? In particular, consider the two limits
a) The CNB neutrinos have a mass of m v = 0.15 eV m v = 0.15 eV m_(v)=0.15eVm_{v}=0.15 \mathrm{eV}mv=0.15eV.
b) The CNB neutrinos have very small masses ( m v / ( k B T ) 0 ) m v / k B T 0 (m_(v)//(k_(B)T)rarr0)\left(m_{v} /\left(k_{B} T\right) \rightarrow 0\right)(mv/(kBT)0).
Hint: In a gas of particles at temperature T T TTT, the mean kinetic energy of the particles is given by E k = 3 k B T / 2 E k = 3 k B T / 2 E_(k)=3k_(B)T//2E_{k}=3 k_{B} T / 2Ek=3kBT/2, where k B 8.6 10 5 eV / K k B 8.6 10 5 eV / K k_(B)≃8.6*10^(-5)eV//Kk_{B} \simeq 8.6 \cdot 10^{-5} \mathrm{eV} / \mathrm{K}kB8.6105eV/K is Boltzmann's constant. The Z 0 Z 0 Z^(0)Z^{0}Z0 mass is m Z 0 91 GeV m Z 0 91 GeV m_(Z^(0))≃91GeVm_{Z^{0}} \simeq 91 \mathrm{GeV}mZ091GeV.
Problem 1.96 Consider elastic scattering of photons on electrons
γ ( k ) + e ( p ) γ ( k ) + e ( p ) γ ( k ) + e ( p ) γ k + e p gamma(k)+e^(-)(p)longrightarrow gamma(k^('))+e^(-)(p^('))\gamma(k)+e^{-}(p) \longrightarrow \gamma\left(k^{\prime}\right)+e^{-}\left(p^{\prime}\right)γ(k)+e(p)γ(k)+e(p)
where k k kkk and p p ppp are the incoming photon and electron 4-momenta and k k k^(')k^{\prime}k and p p p^(')p^{\prime}p the corresponding outgoing 4 -momenta.
a) In the laboratory system, the incoming electron is at rest and the outgoing photon is scattered at an angle θ θ theta\thetaθ with respect to the direction of the incoming photon. Use invariants to derive the so-called "Compton formula," i.e., the difference between the outgoing and incoming photon wavelengths, as a function of θ θ theta\thetaθ, in units c = 1 c = 1 c=1c=1c=1 and = 1 = 1 ℏ=1\hbar=1=1.
b) Derive the angular frequency (energy) of the outgoing photon in the center-of-mass system in terms of the incoming photon angular frequency (energy) in the laboratory system.
Problem 1.97 In Compton scattering γ + e e + γ γ + e e + γ gamma+e longrightarrow e+gamma\gamma+e \longrightarrow e+\gammaγ+ee+γ, photons of a fixed energy ω ω omega\omegaω are scattered against electrons, which can be considered at rest in the laboratory frame. Compute the kinetic energy of the outgoing electron as a function of the scattering angle θ θ theta\thetaθ of the outgoing photon.
Problem 1.98 Inverse Compton scattering occurs when low-energy photons collide with high-energy electrons. Assuming that the photon and electron are originally moving in the same direction, find an expression for the photon energy after the collision as a function of the initial photon energy, the velocity and mass of the electron, and the scattering angle θ θ theta\thetaθ of the photon.
Problem 1.99 An antimuon μ + μ + mu^(+)\mu^{+}μ+decays into a positron e + e + e^(+)e^{+}e+and two neutrinos v e v e v_(e)v_{e}ve and v ¯ μ v ¯ μ bar(v)_(mu)\bar{v}_{\mu}v¯μ. The reaction is
μ + e + + v e + v ¯ μ μ + e + + v e + v ¯ μ mu^(+)longrightarrowe^(+)+v_(e)+ bar(v)_(mu)\mu^{+} \longrightarrow e^{+}+v_{e}+\bar{v}_{\mu}μ+e++ve+v¯μ
Give an expression for the largest possible total energy of the electron neutrino ν e ν e nu_(e)\nu_{e}νe in the rest frame of the antimuon. You may assume that the neutrino masses are negligible compared to lepton masses.
Problem 1.100 A ρ ρ rho\rhoρ-meson with mass m ρ 770 MeV / c 2 m ρ 770 MeV / c 2 m_(rho)≃770MeV//c^(2)m_{\rho} \simeq 770 \mathrm{MeV} / c^{2}mρ770MeV/c2 sometimes decays into a pair of muons ( μ μ mu^(-)\mu^{-}μand μ + μ + mu^(+)\mu^{+}μ+) with mass m μ = m μ + 106 MeV / c 2 m μ = m μ + 106 MeV / c 2 m_(mu^(-))=m_(mu^(+))≃106MeV//c^(2)m_{\mu^{-}}=m_{\mu^{+}} \simeq 106 \mathrm{MeV} / c^{2}mμ=mμ+106MeV/c2 and a photon, γ γ gamma\gammaγ. What is the maximal kinetic energy that the μ + μ + mu^(+)\mu^{+}μ+can have in this decay in the rest frame of the ρ ρ rho\rhoρ-meson?
Problem 1.101 There is a possibility that neutrinos are their own antiparticles. If this is true, then the so-called neutrinoless double beta decay
76 Ge 76 Se + e + e , 76 Ge 76 Se + e + e , ^(76)Gelongrightarrow^(76)Se+e^(-)+e^(-),{ }^{76} \mathrm{Ge} \longrightarrow{ }^{76} \mathrm{Se}+e^{-}+e^{-},76Ge76Se+e+e,
is allowed. Derive expressions for the maximal and minimal possible values of the sum of the kinetic energy of the electrons in the rest frame of 76 Ge 76 Ge ^(76)Ge{ }^{76} \mathrm{Ge}76Ge. Express your answer in terms of the particle masses.
Problem 1.102 At the LHC (Large Hadron Collider), two photons are measured with 4-momenta
(1.13) p 1 = ω 1 ( 1 , 1 , 0 , 0 ) and p 2 = ω 2 ( 1 , cos θ , sin θ , 0 ) (1.13) p 1 = ω 1 ( 1 , 1 , 0 , 0 )  and  p 2 = ω 2 ( 1 , cos θ , sin θ , 0 ) {:(1.13)p_(1)=omega_(1)(1","1","0","0)quad" and "quadp_(2)=omega_(2)(1","cos theta","sin theta","0):}\begin{equation*} p_{1}=\omega_{1}(1,1,0,0) \quad \text { and } \quad p_{2}=\omega_{2}(1, \cos \theta, \sin \theta, 0) \tag{1.13} \end{equation*}(1.13)p1=ω1(1,1,0,0) and p2=ω2(1,cosθ,sinθ,0)
respectively. Assuming that the photon pair results from the decay of a new particle ϕ ϕ phi\phiϕ such that ϕ γ γ ϕ γ γ phi longrightarrow gamma gamma\phi \longrightarrow \gamma \gammaϕγγ, what is the mass of the new particle?
Problem 1.103 In an accelerator, protons are accelerated until they reach a kinetic energy of 8000 MeV and are then made to collide with protons at rest. If the sum of the kinetic energies of two colliding protons (measured in the center-of-mass system) is larger than the rest energy of a proton-antiproton pair, then such a pair can be formed according to the reaction formula
p + p p + p + p + p ¯ p + p p + p + p + p ¯ p+p longrightarrow p+p+p+ bar(p)p+p \longrightarrow p+p+p+\bar{p}p+pp+p+p+p¯
where p p ppp is a proton and p ¯ p ¯ bar(p)\bar{p}p¯ is an antiproton.
Is the energy 8000 MeV sufficient for the reaction to go? The rest mass of the proton is 938 MeV .
Problem 1.104 Protons at rest are bombarded with π π pi\piπ-mesons. How large kinetic energy do the mesons need to have for the reaction
π + p π + + π + n π + p π + + π + n pi^(-)+p longrightarrowpi^(+)+pi^(-)+n\pi^{-}+p \longrightarrow \pi^{+}+\pi^{-}+nπ+pπ++π+n
to take place? The rest mass of the particles are m π = m π + 140 MeV , m p m π = m π + 140 MeV , m p m_(pi^(-))=m_(pi^(+))~~140MeV,m_(p)~~m_{\pi^{-}}=m_{\pi^{+}} \approx 140 \mathrm{MeV}, m_{p} \approxmπ=mπ+140MeV,mp 938 MeV , and m n 940 MeV m n 940 MeV m_(n)~~940MeVm_{n} \approx 940 \mathrm{MeV}mn940MeV.
Problem 1.105 In the CELSIUS ring at the The Svedberg Laboratory in Uppsala, Sweden, one would like to study the reaction
p + d p + p + n + η . p + d p + p + n + η . p+d longrightarrow p+p+n+eta.p+d \longrightarrow p+p+n+\eta .p+dp+p+n+η.
The available kinetic energy of the protons is T p = 700 MeV T p = 700 MeV T_(p)=700MeVT_{p}=700 \mathrm{MeV}Tp=700MeV and the deuterons (d) can be considered to be at rest. The rest masses of the particles are m p m n m p m n m_(p)~~m_(n)m_{p} \approx m_{n}mpmn, m d m p + m n , m n = 940 MeV m d m p + m n , m n = 940 MeV m_(d)~~m_(p)+m_(n),m_(n)=940MeVm_{d} \approx m_{p}+m_{n}, m_{n}=940 \mathrm{MeV}mdmp+mn,mn=940MeV, and m η = 550 MeV m η = 550 MeV m_(eta)=550MeVm_{\eta}=550 \mathrm{MeV}mη=550MeV.
a) Is the reaction possible?
b) If the kinetic energy of the protons in the beam is increased to T p = 1350 MeV T p = 1350 MeV T_(p)=1350MeVT_{p}=1350 \mathrm{MeV}Tp=1350MeV, what is the maximum kinetic energy that the η η eta\etaη can get in the system in which the nucleons are at rest after the reaction, expressed in terms of the rest masses and the kinetic energies?
Problem 1.106 In neutrino detection, the quasi-elastic ( ν μ + X μ + Y ν μ + X μ + Y nu_(mu)+X longrightarrow mu+Y\nu_{\mu}+X \longrightarrow \mu+Yνμ+Xμ+Y, where X X XXX and Y Y YYY are different nuclei) and 1 π ( v μ + X μ + Y + π 0 ) 1 π v μ + X μ + Y + π 0 1pi(v_(mu)+X longrightarrow mu+Y+pi^(0))1 \pi\left(v_{\mu}+X \longrightarrow \mu+Y+\pi^{0}\right)1π(vμ+Xμ+Y+π0) processes are relevant at relatively low energies. Compute the ratio between the neutrino threshold energies for these processes in the rest frame of the nucleus X X XXX. Express your answer in terms of the different particle masses (the neutrino may be considered massless for the purposes of this problem).
Problem 1.107 Consider the particle collision e + e e + e + e + e + e + e e + e + e + e + e^(-)+e^(-)longrightarrowe^(-)+e^(-)+e^(-)+e^(+)e^{-}+e^{-} \longrightarrow e^{-}+e^{-}+e^{-}+e^{+}e+ee+e+e+e+. Compute the necessary total energy of one of the initial electrons in the rest frame of the other for this process to occur. Also, compute the ratio between this energy and the total required energy in the center-of-momentum frame.
Problem 1.108 We can produce neutral kaons in a proton collision through the reaction p + p p + p + K 0 p + p p + p + K 0 p+p longrightarrow p+p+K^(0)p+p \longrightarrow p+p+K^{0}p+pp+p+K0. Find an expression for the threshold kinetic energy of the protons of this reaction when
a) One proton is stationary in the lab frame (find the threshold kinetic energy of the other proton).
b) Both protons have the same kinetic energy (quote the total kinetic energy of both protons).
Problem 1.109 A particle χ χ chi\chiχ hits a stationary proton p p ppp and undergoes inelastic scattering to a new state χ χ chi^(**)\chi^{*}χ while keeping the proton intact. Determine the threshold kinetic energy of χ χ chi\chiχ for this scattering to occur if m χ = m χ + δ > m χ m χ = m χ + δ > m χ m_(chi^(**))=m_(chi)+delta > m_(chi)m_{\chi^{*}}=m_{\chi}+\delta>m_{\chi}mχ=mχ+δ>mχ. Discuss your result in the limit when δ m χ δ m χ delta≪m_(chi)\delta \ll m_{\chi}δmχ.
Problem 1.110 Neutrinos are emitted from core collapse supernovae. If a core collapse supernova occurs at a distance L L LLL from Earth and each neutrino has a total energy E E EEE, how much more time would pass (in the rest frame of the Earth) until the neutrinos reach us if they have a small mass m > 0 m > 0 m > 0m>0m>0 compared to if they were massless ( m = 0 ) ( m = 0 ) (m=0)(m=0)(m=0) ? Give an exact answer as well as a reasonable approximation for when m E m E m≪Em \ll EmE.
Problem 1.111 An elementary particle of charge e ( e e ( e e(ee(ee(e is the elementary charge) is accelerated from rest in a 100 m long straight insulated vacuum cylinder (a linear accelerator) with a constant electric field of 10 4 V / m 10 4 V / m 10^(4)V//m10^{4} \mathrm{~V} / \mathrm{m}104 V/m across the endpoints.
a) What kinetic energy will the particle obtain after the acceleration?
b) How long time does it take for particle to pass through the tube if it starts from rest? Hint: Use the energy as an integration variable.

1.7 Electromagnetism

Problem 1.112 Show by explicit calculation, using the chain rule for derivation and the properties of the Lorentz transformations, that
(1.14) A μ ( x ) = 0 , (1.14) A μ ( x ) = 0 , {:(1.14)◻A^(mu)(x)=0",":}\begin{equation*} \square A^{\mu}(x)=0, \tag{1.14} \end{equation*}(1.14)Aμ(x)=0,
is invariant under Lorentz transformations, i.e., if A μ ( x ) A μ ( x ) A^(mu)(x)A^{\mu}(x)Aμ(x) is a solution to Eq. (1.14), then A μ ( x ) A μ x A^('mu)(x^('))A^{\prime \mu}\left(x^{\prime}\right)Aμ(x) is a solution to the same equation in the primed variables x = Λ x x = Λ x x^(')=Lambda xx^{\prime}=\Lambda xx=Λx, where Λ Λ Lambda\LambdaΛ is a Lorentz transformation.
Problem 1.113 Show that the gauge transformation A μ A μ = A μ + μ ψ A μ A μ = A μ + μ ψ A_(mu)|->A_(mu)^(')=A_(mu)+del_(mu)psiA_{\mu} \mapsto A_{\mu}^{\prime}=A_{\mu}+\partial_{\mu} \psiAμAμ=Aμ+μψ, where ψ ψ psi\psiψ is an arbitrary scalar field, does not affect the field tensor F μ ν = μ A ν ν A μ F μ ν = μ A ν ν A μ F_(mu nu)=del_(mu)A_(nu)-del_(nu)A_(mu)F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}Fμν=μAννAμ.
Problem 1.114 An inertial coordinate system K K K^(')K^{\prime}K is moving relative to another inertial coordinate system K K KKK with constant velocity v v vvv along the positive x 1 x 1 x^(1)x^{1}x1-axis of K K KKK.
a) Assume that a stick of length \ell is at rest in K K KKK such that Δ x = ( , 0 , 0 ) Δ x = ( , 0 , 0 ) Deltax=(ℓ,0,0)\Delta \mathbf{x}=(\ell, 0,0)Δx=(,0,0). Calculate Δ x Δ x Deltax^(')\Delta \mathbf{x}^{\prime}Δx in K K K^(')K^{\prime}K.
b) Assume that there is a constant electric field E = ( 0 , 0 , E ) E = ( 0 , 0 , E ) E=(0,0,E)\mathbf{E}=(0,0, E)E=(0,0,E) in K K KKK (no magnetic field, i.e., B = 0 B = 0 B=0\mathbf{B}=\mathbf{0}B=0 in K K KKK ). Calculate E E E^(')\mathbf{E}^{\prime}E and B B B^(')\mathbf{B}^{\prime}B in K K K^(')K^{\prime}K.
Problem 1.115 An observer at rest in a frame K K KKK experiences only an electric field E E E\mathbf{E}E. Another observer in another frame K K K^(')K^{\prime}K, moving with velocity v v vvv along the positive x x xxx-axis, will observe a magnetic field B B B^(')\mathbf{B}^{\prime}B. Calculate this magnetic field for small velocities (linear terms in v v vvv ) and show that this field is perpendicular to both the electric field E E E^(')\mathbf{E}^{\prime}E and the velocity of K K KKK relative to K K K^(')K^{\prime}K.
Problem 1.116 Let K , K K , K K,K^(')K, K^{\prime}K,K, and K K K^('')K^{\prime \prime}K be as in Problem 1.34. Assume that there is a constant electric field E = ( 0 , 1 , 0 ) E = ( 0 , 1 , 0 ) E=(0,1,0)\mathbf{E}=(0,1,0)E=(0,1,0) (in some given physical units) in the coordinate system K K KKK. We assume that the magnetic field B B B\mathbf{B}B vanishes in K K KKK. Compute the components of both the electric and magnetic fields in the coordinate systems K K K^(')K^{\prime}K and K K K^('')K^{\prime \prime}K.
Problem 1.117 Compute the electric and magnetic field components due to a point charge q q qqq moving with velocity v v vvv along the positive x x xxx-axis.
Problem 1.118 A particle of mass m m mmm and electric charge q q qqq is moving in a constant electric field E E E\boldsymbol{E}E. Use the Lorentz force law to calculate the velocity of the particle as a function of the displacement r r rrr from the origin along the direction of motion. The particle starts off at rest.
Problem 1.119 A current I I III is flowing through a straight uncharged conductor. Determine the electromagnetic field in an inertial system K K K^(')K^{\prime}K that moves parallel to the conductor with velocity v v vvv
a) by transforming the electromagnetic field tensor from the rest frame K K KKK of the conductor to K K K^(')K^{\prime}K,
b) by transforming the current-density 4 -vector from K K KKK to K K K^(')K^{\prime}K, and then, knowing the charge of the conductor and its current relative to K K KKK determine the field in K K K^(')K^{\prime}K.
Problem 1.120 Maxwell's equations can be expressed by means of the electromagnetic 4 -vector potential A A AAA. When μ A μ = 0 μ A μ = 0 del_(mu)A^(mu)=0\partial_{\mu} A^{\mu}=0μAμ=0 (i.e., the Lorenz gauge), they take on a simple form. What is this form? Assuming that Maxwell's equations are in this simple form, and furthermore, J = 0 J = 0 J=0J=0J=0 (i.e., current free), show for a plane wave, A μ = ε μ e i k x A μ = ε μ e i k x A^(mu)=epsi^(mu)e^(ik*x)A^{\mu}=\varepsilon^{\mu} e^{i k \cdot x}Aμ=εμeikx, where ε ε epsi\varepsilonε is the polarization vector, that
(1.15) E k = B k = 0 (1.15) E k = B k = 0 {:(1.15)E*k=B*k=0:}\begin{equation*} \boldsymbol{E} \cdot \boldsymbol{k}=\boldsymbol{B} \cdot \boldsymbol{k}=0 \tag{1.15} \end{equation*}(1.15)Ek=Bk=0
i.e., the electric and magnetic fields are perpendicular to the direction of motion.
Problem 1.121 Calculate the Lorentz invariants F μ ν F μ ν F μ ν F μ ν F_(mu nu)F^(mu nu)F_{\mu \nu} F^{\mu \nu}FμνFμν and ϵ μ ν ω λ F μ ν F ω λ ϵ μ ν ω λ F μ ν F ω λ epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda)\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda}ϵμνωλFμνFωλ for a free electromagnetic plane wave A μ ( x ) = ϵ μ e i k x A μ ( x ) = ϵ μ e i k x A^(mu)(x)=epsilon^(mu)e^(ik*x)A^{\mu}(x)=\epsilon^{\mu} e^{i k \cdot x}Aμ(x)=ϵμeikx, where ϵ ϵ epsilon\epsilonϵ is the polarization vector. Give a physical interpretation of your result.
Problem 1.122 a) Prove that the scalar product E B E B E*B\boldsymbol{E} \cdot \boldsymbol{B}EB between the electric and magnetic field vectors is invariant under Lorentz transformations.
b) Show that if the electric and magnetic fields E E E\boldsymbol{E}E and B B B\boldsymbol{B}B are orthogonal for one observer, they are orthogonal for any observer.
c) Show that E E E\boldsymbol{E}E and B B B\boldsymbol{B}B are orthogonal for free plane waves with A μ ( x ) = ε μ e i k x A μ ( x ) = ε μ e i k x A^(mu)(x)=epsi^(mu)e^(ik*x)A^{\mu}(x)=\varepsilon^{\mu} e^{i k \cdot x}Aμ(x)=εμeikx, where ε ε epsi\varepsilonε is the polarization vector.
d) Show for the plane waves that E × B = A k E × B = A k E xx B=AkE \times B=A kE×B=Ak, where k k kkk is the wave vector and A A AAA is a nonvanishing expression.
Problem 1.123 An electron with mass m 0 m 0 m_(0)m_{0}m0 is moving in a homogeneous magnetic field B = ( 0 , 0 , B ) B = ( 0 , 0 , B ) B=(0,0,B)B=(0,0, B)B=(0,0,B) and no electric field. Calculate its trajectory if it has velocity u = ( u , 0 , 0 ) u = ( u , 0 , 0 ) u=(u,0,0)\boldsymbol{u}=(u, 0,0)u=(u,0,0) at time t = 0 t = 0 t=0t=0t=0.
Problem 1.124 In an inertial coordinate system K K KKK, there is a constant electric field E = ( c B , 0 , 0 ) E = ( c B , 0 , 0 ) E=(cB,0,0)\mathbf{E}=(c B, 0,0)E=(cB,0,0) and a constant magnetic field B = ( 0 , B , 0 ) B = ( 0 , B , 0 ) B=(0,B,0)\mathbf{B}=(0, B, 0)B=(0,B,0). In another inertial system K K K^(')K^{\prime}K, the same fields are measured to be E = ( 0 , 2 c B , c B ) E = ( 0 , 2 c B , c B ) E^(')=(0,2cB,cB)\mathbf{E}^{\prime}=(0,2 c B, c B)E=(0,2cB,cB) and the x x xxx-component B x = B x = B_(x)^(')=B_{x}^{\prime}=Bx= 0 . Compute B y B y B_(y)^(')B_{y}^{\prime}By and B z B z B_(z)^(')B_{z}^{\prime}Bz.
Problem 1.125 Observer A A AAA measures the electric and magnetic field strengths to be E = ( α , α , 0 ) E = ( α , α , 0 ) E=(alpha,-alpha,0)\mathbf{E}=(\alpha,-\alpha, 0)E=(α,α,0) and B = ( 0 , 0 , 2 α / c ) B = ( 0 , 0 , 2 α / c ) B=(0,0,2alpha//c)\mathbf{B}=(0,0,2 \alpha / c)B=(0,0,2α/c), respectively, where α 0 α 0 alpha!=0\alpha \neq 0α0. Another observer, observer B B BBB, makes the same measurements and finds E = ( 0 , 0 , 2 α ) E = ( 0 , 0 , 2 α ) E^(')=(0,0,2alpha)\mathbf{E}^{\prime}=(0,0,2 \alpha)E=(0,0,2α) and B = ( B x , α / c , B z ) B = B x , α / c , B z B^(')=(B_(x)^('),alpha//c,B_(z)^('))\mathbf{B}^{\prime}=\left(B_{x}^{\prime}, \alpha / c, B_{z}^{\prime}\right)B=(Bx,α/c,Bz). Determine B x B x B_(x)^(')B_{x}^{\prime}Bx and B z B z B_(z)^(')B_{z}^{\prime}Bz.
Problem 1.126 Observer A A AAA measures the electric and magnetic field strengths to be E = ( 0 , β , β ) E = ( 0 , β , β ) E=(0,beta,-beta)\mathbf{E}=(0, \beta,-\beta)E=(0,β,β) and B = ( 2 β / c , 0 , 0 ) B = ( 2 β / c , 0 , 0 ) B=(2beta//c,0,0)\mathbf{B}=(2 \beta / c, 0,0)B=(2β/c,0,0), respectively, where β 0 β 0 beta!=0\beta \neq 0β0. Another observer, observer B B BBB, makes the same measurements and finds E = ( 2 β , 0 , 0 ) E = ( 2 β , 0 , 0 ) E^(')=(2beta,0,0)\mathbf{E}^{\prime}=(2 \beta, 0,0)E=(2β,0,0) and B = ( B x , B y , β / c ) B = B x , B y , β / c B^(')=(B_(x)^('),B_(y)^('),beta//c)\mathbf{B}^{\prime}=\left(B_{x}^{\prime}, B_{y}^{\prime}, \beta / c\right)B=(Bx,By,β/c). Determine B x B x B_(x)^(')B_{x}^{\prime}Bx and B y B y B_(y)^(')B_{y}^{\prime}By.
Problem 1.127 Observer A A AAA measures the electric and magnetic field strengths to be E = ( α , 0 , 0 ) E = ( α , 0 , 0 ) E=(alpha,0,0)\mathbf{E}=(\alpha, 0,0)E=(α,0,0) and B = ( α / c , 0 , 2 α / c ) B = ( α / c , 0 , 2 α / c ) B=(alpha//c,0,2alpha//c)\mathbf{B}=(\alpha / c, 0,2 \alpha / c)B=(α/c,0,2α/c), respectively, where α 0 α 0 alpha!=0\alpha \neq 0α0. Another observer, observer B B BBB, makes the same measurements and finds E = ( E x , α , 0 ) E = E x , α , 0 E^(')=(E_(x)^('),alpha,0)\mathbf{E}^{\prime}=\left(E_{x}^{\prime}, \alpha, 0\right)E=(Ex,α,0) and B = ( α / c , B y , α / c ) B = α / c , B y , α / c B^(')=(alpha//c,B_(y)^('),alpha//c)\mathbf{B}^{\prime}=\left(\alpha / c, B_{y}^{\prime}, \alpha / c\right)B=(α/c,By,α/c). Express E x E x E_(x)^(')E_{x}^{\prime}Ex and B y B y B_(y)^(')B_{y}^{\prime}By in terms of α α alpha\alphaα and c c ccc. Finally, a third observer, observer C C CCC, is moving relative to observer B B BBB with constant velocity v v vvv along the positive x x xxx-axis of observer B B BBB. Find the electric and magnetic field strengths, E E E^('')\mathbf{E}^{\prime \prime}E and B B B^('')\mathbf{B}^{\prime \prime}B, as observer C C CCC measures them.
Problem 1.128 Assume that a muon originally travels vertically down toward the ground from an altitude of 10 km . There is a magnetic field coming from the Earth of B = 50 μ T B = 50 μ T B=50 muTB=50 \mu \mathrm{~T}B=50μ T affecting the motion of the muon. To make a simple model we take the magnetic field to be constant all the way from 10 km altitude to ground level. Suppose the field lines go from south to north and we are in Japan on the northern hemisphere. How far in length and in which direction is the deviation from the point where the muon would hit the ground without magnetic field, compared to where it hits the ground due to the deviation induced by the magnetic field of the Earth, if it has the energy of 2 GeV and is negatively charged?
Hint: The combination c B c B cBc BcB, where c c ccc is the speed of light, has the value c B = 300 V / m c B = 300 V / m cB=300V//mc B=300 \mathrm{~V} / \mathrm{m}cB=300 V/m, for B = 1 μ T B = 1 μ T B=1muTB=1 \mu \mathrm{~T}B=1μ T. The trajectory of a charged particle in a homogeneous magnetic field is a circle, it is sufficient to compute the radius of the circle and then use geometric arguments.
Problem 1.129 An observer in the system S S SSS has observed an electromagnetic field tensor F μ ν F μ ν F^(mu nu)F^{\mu \nu}Fμν with nonvanishing E E E\boldsymbol{E}E - and B B B\boldsymbol{B}B-fields. Performing a Lorentz transformation with velocity u u uuu along the positive x 1 x 1 x_(1)x_{1}x1-axis to another system S S S^(')S^{\prime}S he finds that the B B BBB-field is absent, i.e., all its components are equal to 0 . What is the electric field in this system expressed in u u uuu and the components of the electric field in S S SSS ?
Problem 1.130 In an inertial frame S S SSS there is a constant time-independent magnetic field B B BBB and no electric field ( E = 0 ) ( E = 0 ) (E=0)(\boldsymbol{E}=\mathbf{0})(E=0). Consider another inertial frame S S S^(')S^{\prime}S, which moves with velocity v v vvv along the positive x 1 x 1 x^(1)x^{1}x1-axis of S S SSS.
a) What are the E E E^(')\boldsymbol{E}^{\prime}E and B B B^(')\boldsymbol{B}^{\prime}B fields in the system S S S^(')S^{\prime}S expressed in the original B B B\boldsymbol{B}B-field and the velocity v v vvv ?
b) Verify that the Lorentz invariants are indeed invariant under this transformation.
Problem 1.131 An electron in a linear particle accelerator of length L = 3 km L = 3 km L=3kmL=3 \mathrm{~km}L=3 km (e.g., SLAC in California, USA) is accelerated through an electric potential U U UUU.
a) Compute the trajectory x ( t ) x ( t ) x(t)x(t)x(t) of this electron for 0 < | x ( t ) | < L 0 < | x ( t ) | < L 0 < |x(t)| < L0<|x(t)|<L0<|x(t)|<L if its motion starts at time t = 0 t = 0 t=0t=0t=0 at rest at one end of the accelerator.
b) Compute the time it takes for this electron to pass through the whole accelerator.
c) Compute the time dependence of the energy of this electron in the accelerator.
Problem 1.132 a) Find the electric and magnetic fields E E E\boldsymbol{E}E and B B BBB generated by a particle with charge q q qqq moving with constant velocity v v vvv parallel with the x x xxx-axis in an
inertial system S S SSS, using that the electric and magnetic potentials in the particle's rest frame are
(1.16) ϕ ( t , x ) = q 4 π | x | , A ( t , x ) = 0 (1.16) ϕ t , x = q 4 π x , A t , x = 0 {:(1.16)phi(t^('),x^('))=(q)/(4pi|x^(')|)","quad A(t^('),x^('))=0:}\begin{equation*} \phi\left(t^{\prime}, \boldsymbol{x}^{\prime}\right)=\frac{q}{4 \pi\left|\boldsymbol{x}^{\prime}\right|}, \quad \boldsymbol{A}\left(t^{\prime}, \boldsymbol{x}^{\prime}\right)=0 \tag{1.16} \end{equation*}(1.16)ϕ(t,x)=q4π|x|,A(t,x)=0
we use the notation A = ( A 1 , A 2 , A 3 ) A = A 1 , A 2 , A 3 A=(A^(1),A^(2),A^(3))\boldsymbol{A}=\left(A^{1}, A^{2}, A^{3}\right)A=(A1,A2,A3), and similarly for E , B E , B E,B\boldsymbol{E}, \boldsymbol{B}E,B, and x x x\boldsymbol{x}x.
b) Explain why it is possible to check your result in a) by computing E B E B E*B\boldsymbol{E} \cdot \boldsymbol{B}EB and E 2 B 2 E 2 B 2 E^(2)-B^(2)\boldsymbol{E}^{2}-\boldsymbol{B}^{2}E2B2 in both inertial systems. Perform these checks!
Problem 1.133 Bubble chambers were frequently used in the 1960s in particle collision experiments. In a bubble chamber, there is a strong constant magnetic field, which bends the motion of charged particles. The charged particles give rise to bubbles, which make the trajectories of the charged particles visible.
a) In the lab frame of the bubble chamber, there is a strong magnetic field in the z z zzz-direction and no electric field. Use the Lorentz force law to show that the trajectory of a charge particle can be parametrized in the lab frame as
(1.17) x = R cos ω τ , y = R sin ω τ , (1.17) x = R cos ω τ , y = R sin ω τ , {:(1.17)x=R cos omega tau","quad y=-R sin omega tau",":}\begin{equation*} x=R \cos \omega \tau, \quad y=-R \sin \omega \tau, \tag{1.17} \end{equation*}(1.17)x=Rcosωτ,y=Rsinωτ,
and determine ω ω omega\omegaω. Show that for a charged particle, you can obtain the 3-momentum from knowing the radius of the trajectory and the strength of the magnetic field (any energy losses can be neglect)
(1.18) p i p i = q 2 R 2 B i B i (1.18) p i p i = q 2 R 2 B i B i {:(1.18)p_(i)p^(i)=q^(2)R^(2)B_(i)B^(i):}\begin{equation*} p_{i} p^{i}=q^{2} R^{2} B_{i} B^{i} \tag{1.18} \end{equation*}(1.18)pipi=q2R2BiBi
b) In a bubble chamber, one can only see the traces of charged particles in terms of bubbles. Consider the following process
Σ π + X 0 Σ π + X 0 Sigma^(-)longrightarrowpi^(-)+X^(0)\Sigma^{-} \longrightarrow \pi^{-}+X^{0}Σπ+X0
where Σ Σ Sigma^(-)\Sigma^{-}Σand π π pi^(-)\pi^{-}πare known charged particles. Here X 0 X 0 X^(0)X^{0}X0 is an unknown uncharged particle, which we cannot see, since it does not give rise to bubbles. For the other two particles, we know their rest masses and their trajectory radii R Σ R Σ R_(Sigma)R_{\Sigma}RΣ and R π R π R_(pi)R_{\pi}Rπ (therefore, we also know their 3-momenta). From this information, derive an expression for the rest mass of the unknown particle expressed in terms of the rest masses M Σ M Σ M_(Sigma)M_{\Sigma}MΣ of Σ Σ Sigma^(-)\Sigma^{-}Σ and M π M π M_(pi)M_{\pi}Mπ of π π pi^(-)\pi^{-}π, and their respective 3-momenta, as well as the angle θ θ theta\thetaθ between the recorded trajectories of the charged particles close to the collision.
Problem 1.134 Starting from the plane wave solution to Maxwell's equations
(1.19) A μ = ε μ sin ( k x ) (1.19) A μ = ε μ sin ( k x ) {:(1.19)A^(mu)=epsi^(mu)sin(k*x):}\begin{equation*} A^{\mu}=\varepsilon^{\mu} \sin (k \cdot x) \tag{1.19} \end{equation*}(1.19)Aμ=εμsin(kx)
show that the electric and magnetic fields are orthogonal and have the same magnitude without referring to a particular gauge condition.
Problem 1.135 Assume that the electromagnetic field in an inertial frame S S SSS satisfies | E | = | B | | E | = | B | |E|=|B||\boldsymbol{E}|=|\boldsymbol{B}||E|=|B| and that the angle between the electric and magnetic field is α α alpha\alphaα. In another inertial frame, the fields are E E E^(')\boldsymbol{E}^{\prime}E and B B B^(')\boldsymbol{B}^{\prime}B with a corresponding angle α α alpha^(')\alpha^{\prime}α. Show that
(1.20) cos α = E 2 E 2 cos α (1.20) cos α = E 2 E 2 cos α {:(1.20)cos alpha^(')=(E^(2))/(E^('2))*cos alpha:}\begin{equation*} \cos \alpha^{\prime}=\frac{\boldsymbol{E}^{2}}{\boldsymbol{E}^{\prime 2}} \cdot \cos \alpha \tag{1.20} \end{equation*}(1.20)cosα=E2E2cosα
Problem 1.136 The 4-potential of a stationary point charge Q Q QQQ in its rest frame is given by
(1.21) A μ = Q 4 π r ( 1 , 0 ) μ (1.21) A μ = Q 4 π r ( 1 , 0 ) μ {:(1.21)A^(mu)=(Q)/(4pi r)(1","0)^(mu):}\begin{equation*} A^{\mu}=\frac{Q}{4 \pi r}(1,0)^{\mu} \tag{1.21} \end{equation*}(1.21)Aμ=Q4πr(1,0)μ
where r = x 2 + y 2 + z 2 r = x 2 + y 2 + z 2 r=sqrt(x^(2)+y^(2)+z^(2))r=\sqrt{x^{2}+y^{2}+z^{2}}r=x2+y2+z2 is the distance to the particle. Compute the electromagnetic stress-energy tensor T μ v T μ v T^(mu v)T^{\mu v}Tμv in ( x , y , z ) = ( 1 , 0 , 0 ) ( x , y , z ) = ( 1 , 0 , 0 ) (x,y,z)=(1,0,0)(x, y, z)=(1,0,0)(x,y,z)=(1,0,0) and the corresponding trace T μ μ T μ μ T_(mu)^(mu)T_{\mu}^{\mu}Tμμ.
Problem 1.137 Starting from Maxwell's equations and without assuming a particular gauge condition, show that the components of the electromagnetic field tensor F μ v F μ v F^(mu v)F^{\mu v}Fμv satisfy the sourced wave equation
(1.22) F μ v σ σ F μ v = S μ v (1.22) F μ v σ σ F μ v = S μ v {:(1.22)◻F^(mu v)-=del_(sigma)del^(sigma)F^(mu v)=S^(mu v):}\begin{equation*} \square F^{\mu v} \equiv \partial_{\sigma} \partial^{\sigma} F^{\mu v}=S^{\mu v} \tag{1.22} \end{equation*}(1.22)FμvσσFμv=Sμv
and express the source tensor S μ ν S μ ν S^(mu nu)S^{\mu \nu}Sμν in terms of the 4-current density J μ J μ J^(mu)J^{\mu}Jμ.
Problem 1.138 The electromagnetic stress-energy tensor is given by
(1.23) T μ v = ε 0 [ F μ σ F v σ 1 4 δ μ v ( F ρ σ F ρ σ ) ] (1.23) T μ v = ε 0 F μ σ F v σ 1 4 δ μ v F ρ σ F ρ σ {:(1.23)T_(mu)^(v)=-epsi_(0)[F_(mu sigma)F^(v sigma)-(1)/(4)delta_(mu)^(v)(F_(rho sigma)F^(rho sigma))]:}\begin{equation*} T_{\mu}^{v}=-\varepsilon_{0}\left[F_{\mu \sigma} F^{v \sigma}-\frac{1}{4} \delta_{\mu}^{v}\left(F_{\rho \sigma} F^{\rho \sigma}\right)\right] \tag{1.23} \end{equation*}(1.23)Tμv=ε0[FμσFvσ14δμv(FρσFρσ)]
Given the electromagnetic plane wave solution for the 4 -potential
(1.24) A μ = ε μ sin ( k x ) (1.24) A μ = ε μ sin ( k x ) {:(1.24)A^(mu)=epsi^(mu)sin(k*x):}\begin{equation*} A^{\mu}=\varepsilon^{\mu} \sin (k \cdot x) \tag{1.24} \end{equation*}(1.24)Aμ=εμsin(kx)
express T μ v T μ v T_(mu)^(v)T_{\mu}^{v}Tμv in terms of the 4 -vector k k kkk. You also need to assure that the wave actually fulfills Maxwell's equations in the absence of a source term μ F μ ν = 0 μ F μ ν = 0 del_(mu)F^(mu nu)=0\partial_{\mu} F^{\mu \nu}=0μFμν=0.
Hint: You may assume the Lorenz gauge condition μ A μ = 0 μ A μ = 0 del_(mu)A^(mu)=0\partial_{\mu} A^{\mu}=0μAμ=0.
Problem 1.139 The 4-potential A μ A μ A_(mu)A_{\mu}Aμ is not physical, but may be transformed according to A μ A μ + μ φ A μ A μ + μ φ A_(mu)|->A_(mu)+del_(mu)varphiA_{\mu} \mapsto A_{\mu}+\partial_{\mu} \varphiAμAμ+μφ, where φ φ varphi\varphiφ is a scalar field, without changing the physical observables. Show that the physical electromagnetic field tensor F μ ν F μ ν F^(mu nu)F^{\mu \nu}Fμν is invariant under this transformation.
Problem 1.140 The electric field of an electric dipole with dipole moment d = d e z d = d e z d=de_(z)d=d e_{z}d=dez is given by
(1.25) E = d 4 π ε 0 ( 3 x z r 5 e x + 3 y z r 5 e y + 3 z 2 r 2 r 5 e z ) (1.25) E = d 4 π ε 0 3 x z r 5 e x + 3 y z r 5 e y + 3 z 2 r 2 r 5 e z {:(1.25)E=(d)/(4piepsi_(0))((3xz)/(r^(5))e_(x)+(3yz)/(r^(5))e_(y)+(3z^(2)-r^(2))/(r^(5))e_(z)):}\begin{equation*} \boldsymbol{E}=\frac{d}{4 \pi \varepsilon_{0}}\left(\frac{3 x z}{r^{5}} \boldsymbol{e}_{x}+\frac{3 y z}{r^{5}} \boldsymbol{e}_{y}+\frac{3 z^{2}-r^{2}}{r^{5}} \boldsymbol{e}_{z}\right) \tag{1.25} \end{equation*}(1.25)E=d4πε0(3xzr5ex+3yzr5ey+3z2r2r5ez)
in its rest frame S S SSS. Compute the value of the quantity F μ ν F ~ μ ν = ε μ ν σ ρ F μ ν F σ ρ F μ ν F ~ μ ν = ε μ ν σ ρ F μ ν F σ ρ F_(mu nu) tilde(F)^(mu nu)=epsi^(mu nu sigma rho)F_(mu nu)F_(sigma rho)F_{\mu \nu} \tilde{F}^{\mu \nu}=\varepsilon^{\mu \nu \sigma \rho} F_{\mu \nu} F_{\sigma \rho}FμνF~μν=εμνσρFμνFσρ as a function of time and position in the frame S S S^(')S^{\prime}S, which is moving in the positive x x xxx-direction with velocity v v vvv relative to S S S^(')S^{\prime}S.
Problem 1.141 A particle at rest acting as an electric monopole and a magnetic dipole has the electromagnetic fields
(1.26) E = q 4 π r 2 e r and B = 1 4 π r 3 [ 3 ( m e r ) e r m ] (1.26) E = q 4 π r 2 e r  and  B = 1 4 π r 3 3 m e r e r m {:(1.26)E=(q)/(4pir^(2))e_(r)quad" and "quad B=(1)/(4pir^(3))[3(m*e_(r))e_(r)-m]:}\begin{equation*} E=\frac{q}{4 \pi r^{2}} e_{r} \quad \text { and } \quad B=\frac{1}{4 \pi r^{3}}\left[3\left(m \cdot e_{r}\right) e_{r}-m\right] \tag{1.26} \end{equation*}(1.26)E=q4πr2er and B=14πr3[3(mer)erm]
Figure 1.10 Two particles (each with charge q q qqq ) and the plane S S SSS equidistant from both particles.
where r r rrr is the position vector relative to the particle, r r rrr its magnitude, q q qqq the charge of the particle, and m m mmm its magnetic dipole moment. Determine whether or not there exists a region of spacetime where the electric field is equal to zero in some inertial frame (although that frame may generally be different for different points in the region). If such a region exists, determine the shape of the region in the particle's rest frame.
Problem 1.142 Two particles with the same charge q q qqq are held fixed with a separation distance d d ddd (see Figure 1.10). Compute the stress-energy tensor of the static electric field between the charges and use your result to find the total 4 -force between the electromagnetic fields on either side of the plane S S SSS that is equidistant from both charges.
Problem 1.143 Compute the Lorentz 4 -force between two electrons moving in parallel with constant velocity v v v\boldsymbol{v}v and a separation d d ddd orthogonal to the direction of motion.

1.8 Energy-Momentum Tensor

Problem 1.144 Determine the momentum density of a gas consisting of massless noninteracting particles in a frame which is moving with velocity v v v\boldsymbol{v}v relative to the gas rest frame. Express your result in terms of v v vvv and the energy density of the gas in the frame where the gas is moving.
Problem 1.145 A star cruiser is moving through space with velocity v v vvv relative to the galaxy. Suddenly it encounters a gas cloud of dust particles. What is the 4 -force from the dust cloud on the star cruiser at the moment it enters the cloud? You may assume that the star cruiser has a cross sectional area A A AAA relative to the direction of
motion and that all of the dust particles encountered will be absorbed in the hull of the star cruiser. In addition to computing the 4 -force, motivate and state whether it is pure, heatlike, or neither.
Problem 1.146 The energy-momentum tensor of a string with tension t t ttt is given by
(1.27) ( T μ v ) = ( ρ 0 0 0 σ ) , (1.27) T μ v = ρ 0 0 0 σ , {:(1.27)(T^(mu v))=([rho_(0),0],[0,-sigma])",":}\left(T^{\mu v}\right)=\left(\begin{array}{cc} \rho_{0} & 0 \tag{1.27}\\ 0 & -\sigma \end{array}\right),(1.27)(Tμv)=(ρ000σ),
where ρ 0 ρ 0 rho_(0)\rho_{0}ρ0 is the string density and σ = t / A < ρ 0 σ = t / A < ρ 0 sigma=t//A < rho_(0)\sigma=t / A<\rho_{0}σ=t/A<ρ0 is the stress across the string cross section A A AAA.
a) Does a frame exist where the stress ( T 11 ) T 11 (T^(11))\left(T^{11}\right)(T11) is equal to zero?
b) Does a frame exist where the energy density is smaller than ρ 0 ρ 0 rho_(0)\rho_{0}ρ0 ?
Your answers should be accompanied by solid argumentation.
Problem 1.147 A pure photon gas such as the cosmic microwave background (CMB) can be described as a perfect fluid with pressure p = ρ 0 / 3 p = ρ 0 / 3 p=rho_(0)//3p=\rho_{0} / 3p=ρ0/3 in its rest frame. In a frame moving with velocity v v vvv in relation to the rest frame of the CMB, compute the energy density, momentum density, and stress tensors. In addition, comment on whether the shear stress (off-diagonal elements of the stress tensor) in an arbitrary frame is zero or not.
Problem 1.148 In a perfect fluid with proper density ρ 0 ρ 0 rho_(0)\rho_{0}ρ0 and positive proper pressure p p ppp, find an expression for the energy density ρ ρ rho\rhoρ in an arbitrary inertial frame S S S^(')S^{\prime}S and derive an upper bound on ρ / γ 2 ρ / γ 2 rho//gamma^(2)\rho / \gamma^{2}ρ/γ2, where γ γ gamma\gammaγ is the gamma factor between the fluid's rest frame and S S S^(')S^{\prime}S.
Problem 1.149 The energy density in the frame of an observer with 4-velocity V V VVV is given by ρ = T μ ν V μ V ν ρ = T μ ν V μ V ν rho=T_(mu nu)V^(mu)V^(nu)\rho=T_{\mu \nu} V^{\mu} V^{\nu}ρ=TμνVμVν. The weak energy condition is a condition requiring the energy density to be nonnegative for all observers, i.e., ρ 0 ρ 0 rho >= 0\rho \geq 0ρ0. For a perfect fluid, determine the condition on the equation of state parameter w w www in the relation p = w ρ 0 p = w ρ 0 p=wrho_(0)p=w \rho_{0}p=wρ0 that the weak energy condition implies.

1.9 Lagrange's Formalism

Problem 1.150 The 4-momentum of a free particle of mass m m mmm is p μ = m c x ˙ μ p μ = m c x ˙ μ p^(mu)=mcx^(˙)^(mu)p^{\mu}=m c \dot{x}^{\mu}pμ=mcx˙μ.
a) Show that the momentum is conserved (i.e., independent of time) by deriving the Euler-Lagrange variational equations for the Lagrangian L = p 2 / ( 2 m ) L = p 2 / ( 2 m ) L=p^(2)//(2m)\mathscr{L}=p^{2} /(2 m)L=p2/(2m) in Minkowski space.
b) When the particle moves in an electromagnetic field, one can obtain the relevant equations of motion by using the substitution p p + q A / c p p + q A / c p|->p+qA//cp \mapsto p+q A / cpp+qA/c, where A = A ( x ) A = A ( x ) A=A(x)A=A(x)A=A(x) is the electromagnetic potential and q q qqq is the charge of the particle. Show that, to lowest nontrivial order in q q qqq, the equations of motion for the particle give the equations of the Lorentz force.

General Relativity Theory

2.1 Some Differential Geometry

Problem 2.1 Show that the function f ( x , y ) = x 2 + y f ( x , y ) = x 2 + y f(x,y)=x^(2)+yf(x, y)=x^{2}+yf(x,y)=x2+y is a smooth function on the unit sphere S 2 R 3 S 2 R 3 S^(2)subR^(3)\mathbb{S}^{2} \subset \mathbb{R}^{3}S2R3.
Problem 2.2 On the unit sphere M = S 2 M = S 2 M=S^(2)M=\mathbb{S}^{2}M=S2, we use the spherical coordinates θ θ theta\thetaθ and ϕ ϕ phi\phiϕ, except at the poles θ = 0 , π θ = 0 , π theta=0,pi\theta=0, \piθ=0,π. A curve can then be parametrized as ( θ ( s ) , ϕ ( s ) ) ( θ ( s ) , ϕ ( s ) ) (theta(s),phi(s))(\theta(s), \phi(s))(θ(s),ϕ(s)). A tangent vector v T p S 2 v T p S 2 v inT_(p)S^(2)v \in T_{p} \mathbb{S}^{2}vTpS2 is given by its components v = ( v θ , v ϕ ) v = v θ , v ϕ v=(v_(theta),v_(phi))v=\left(v_{\theta}, v_{\phi}\right)v=(vθ,vϕ) with v θ = θ ˙ ( s 0 ) v θ = θ ˙ s 0 v_(theta)=theta^(˙)(s_(0))v_{\theta}=\dot{\theta}\left(s_{0}\right)vθ=θ˙(s0), v ϕ = ϕ ( s 0 ) v ϕ = ϕ s 0 v_(phi)=phi(s_(0))v_{\phi}=\phi\left(s_{0}\right)vϕ=ϕ(s0), and p = ( θ ( s 0 ) , ϕ ( s 0 ) ) p = θ s 0 , ϕ s 0 p=(theta(s_(0)),phi(s_(0)))p=\left(\theta\left(s_{0}\right), \phi\left(s_{0}\right)\right)p=(θ(s0),ϕ(s0)). How would you describe a tangent vector at the poles θ = 0 , π θ = 0 , π theta=0,pi\theta=0, \piθ=0,π ?
Problem 2.3 Find the metric tensor and the Christoffel symbols in the twodimensional Euclidean plane in the following coordinates
a) s s sss and t t ttt defined by x = s e t x = s e t x=se^(t)x=s e^{t}x=set and y = s e t y = s e t y=se^(-t)y=s e^{-t}y=set.
b) u u uuu and v v vvv defined by x = u x = u x=ux=ux=u and y = v 2 y = v 2 y=v^(2)y=v^{2}y=v2.
In both cases, discuss where in the Euclidean plane the new coordinates provide a well-defined coordinate system.
Problem 2.4 Let α ( t ) α ( t ) alpha(t)\alpha(t)α(t) and β ( t ) β ( t ) beta(t)\beta(t)β(t) be a pair of smooth curves on a manifold M M MMM such that α ( t 0 ) = β ( t 0 ) α t 0 = β t 0 alpha(t_(0))=beta(t_(0))\alpha\left(t_{0}\right)=\beta\left(t_{0}\right)α(t0)=β(t0). Show that the condition
(2.1) d d t x i ( α ( t ) ) | t = t 0 = d d t x i ( β ( t ) ) | t = t 0 for i = 1 , 2 , , n , (2.1) d d t x i ( α ( t ) ) t = t 0 = d d t x i ( β ( t ) ) t = t 0  for  i = 1 , 2 , , n , {:(2.1)(d)/(dt)x^(i)(alpha(t))|_(t=t_(0))=(d)/(dt)x^(i)(beta(t))|_(t=t_(0))quad" for "i=1","2","dots","n",":}\begin{equation*} \left.\frac{d}{d t} x^{i}(\alpha(t))\right|_{t=t_{0}}=\left.\frac{d}{d t} x^{i}(\beta(t))\right|_{t=t_{0}} \quad \text { for } i=1,2, \ldots, n, \tag{2.1} \end{equation*}(2.1)ddtxi(α(t))|t=t0=ddtxi(β(t))|t=t0 for i=1,2,,n,
is independent of the choice of local coordinates x i x i x^(i)x^{i}xi, i.e., if the curves are tangential in one coordinate system, then they are tangential in any other coordinate system.
Problem 2.5 Show that the system of first-order ordinary differential equations
(2.2) x ˙ k k Y i = Y ˙ i ( s ) + Γ k j i ( x ( s ) ) x ˙ k ( s ) Y j ( s ) = 0 , i = 1 , 2 , , n , (2.2) x ˙ k k Y i = Y ˙ i ( s ) + Γ k j i ( x ( s ) ) x ˙ k ( s ) Y j ( s ) = 0 , i = 1 , 2 , , n , {:(2.2)x^(˙)^(k)grad_(k)Y^(i)=Y^(˙)^(i)(s)+Gamma_(kj)^(i)(x(s))x^(˙)^(k)(s)Y^(j)(s)=0","quad i=1","2","dots","n",":}\begin{equation*} \dot{x}^{k} \nabla_{k} Y^{i}=\dot{Y}^{i}(s)+\Gamma_{k j}^{i}(x(s)) \dot{x}^{k}(s) Y^{j}(s)=0, \quad i=1,2, \ldots, n, \tag{2.2} \end{equation*}(2.2)x˙kkYi=Y˙i(s)+Γkji(x(s))x˙k(s)Yj(s)=0,i=1,2,,n,
defining the parallel transport along a curve on a manifold M M MMM is coordinate independent in the sense that if the system is valid in one coordinate system, then it is also valid in any other coordinate system.
Problem 2.6 The distance between two points a a aaa and b b bbb on the unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2 along a curve γ ( s ) = ( θ ( s ) , ϕ ( s ) ) γ ( s ) = ( θ ( s ) , ϕ ( s ) ) gamma(s)=(theta(s),phi(s))\gamma(s)=(\theta(s), \phi(s))γ(s)=(θ(s),ϕ(s)) is given by
[ γ ] a b g γ ( s ) ( γ ˙ ( s ) , γ ˙ ( s ) ) d s = a b g θ θ θ ˙ ( s ) 2 + g ϕ ϕ ϕ ˙ ( s ) 2 d s (2.3) = a b θ ˙ ( s ) 2 + sin 2 θ ( s ) ϕ ˙ ( s ) 2 d s [ γ ] a b g γ ( s ) ( γ ˙ ( s ) , γ ˙ ( s ) ) d s = a b g θ θ θ ˙ ( s ) 2 + g ϕ ϕ ϕ ˙ ( s ) 2 d s (2.3) = a b θ ˙ ( s ) 2 + sin 2 θ ( s ) ϕ ˙ ( s ) 2 d s {:[ℓ[gamma]-=int_(a)^(b)sqrt(g_(gamma(s))(gamma^(˙)(s),gamma^(˙)(s)))ds=int_(a)^(b)sqrt(g_(theta theta)theta^(˙)(s)^(2)+g_(phi phi)phi^(˙)(s)^(2))ds],[(2.3)=int_(a)^(b)sqrt(theta^(˙)(s)^(2)+sin^(2)theta(s)phi^(˙)(s)^(2))ds]:}\begin{align*} \ell[\gamma] & \equiv \int_{a}^{b} \sqrt{g_{\gamma(s)}(\dot{\gamma}(s), \dot{\gamma}(s))} d s=\int_{a}^{b} \sqrt{g_{\theta \theta} \dot{\theta}(s)^{2}+g_{\phi \phi} \dot{\phi}(s)^{2}} d s \\ & =\int_{a}^{b} \sqrt{\dot{\theta}(s)^{2}+\sin ^{2} \theta(s) \dot{\phi}(s)^{2}} d s \tag{2.3} \end{align*}[γ]abgγ(s)(γ˙(s),γ˙(s))ds=abgθθθ˙(s)2+gϕϕϕ˙(s)2ds(2.3)=abθ˙(s)2+sin2θ(s)ϕ˙(s)2ds
Use Euler-Lagrange equations to derive the geodesic equations on S 2 S 2 S^(2)\mathbb{S}^{2}S2.
Problem 2.7 Compute the Christoffel symbols on the unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2 with metric given by d s 2 = d θ 2 + sin 2 θ d ϕ 2 d s 2 = d θ 2 + sin 2 θ d ϕ 2 ds^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d s^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}ds2=dθ2+sin2θdϕ2
a) directly from the metric.
b) using the general formula for the geodesic equations.
Problem 2.8 We define the Christoffel symbols on the unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2, using spherical coordinates ( θ , ϕ ) ( θ , ϕ ) (theta,phi)(\theta, \phi)(θ,ϕ). When θ 0 , π θ 0 , π theta!=0,pi\theta \neq 0, \piθ0,π, we find (see Problem 2.7)
(2.4) Γ ϕ ϕ θ = 1 2 sin 2 θ , Γ θ ϕ ϕ = Γ ϕ θ ϕ = cot θ , (2.4) Γ ϕ ϕ θ = 1 2 sin 2 θ , Γ θ ϕ ϕ = Γ ϕ θ ϕ = cot θ , {:(2.4)Gamma_(phi phi)^(theta)=-(1)/(2)sin 2theta","quadGamma_(theta phi)^(phi)=Gamma_(phi theta)^(phi)=cot theta",":}\begin{equation*} \Gamma_{\phi \phi}^{\theta}=-\frac{1}{2} \sin 2 \theta, \quad \Gamma_{\theta \phi}^{\phi}=\Gamma_{\phi \theta}^{\phi}=\cot \theta, \tag{2.4} \end{equation*}(2.4)Γϕϕθ=12sin2θ,Γθϕϕ=Γϕθϕ=cotθ,
and all other Γ Γ Gamma\GammaΓ are equal to zero. Show that the apparent singularity at θ = 0 , π θ = 0 , π theta=0,pi\theta=0, \piθ=0,π can be removed by a better choice of coordinates at the poles of the sphere. Thus, the above affine connection extends to the whole S 2 S 2 S^(2)\mathbb{S}^{2}S2.
Problem 2.9 a) Let M = S 2 M = S 2 M=S^(2)M=\mathbb{S}^{2}M=S2 and Γ Γ Gamma\GammaΓ be the affine connection in Problem 2.8. The coordinates θ ( s ) θ ( s ) theta(s)\theta(s)θ(s) and ϕ ( s ) ϕ ( s ) phi(s)\phi(s)ϕ(s) of a geodesic then satisfy the geodesic equations, i.e.,
(2.5) θ ¨ ( s ) 1 2 sin 2 θ ( s ) ϕ ˙ ( s ) ϕ ˙ ( s ) = 0 (2.6) ϕ ¨ ( s ) + 2 cot θ ( s ) θ ˙ ( s ) ϕ ˙ ( s ) = 0 (2.5) θ ¨ ( s ) 1 2 sin 2 θ ( s ) ϕ ˙ ( s ) ϕ ˙ ( s ) = 0 (2.6) ϕ ¨ ( s ) + 2 cot θ ( s ) θ ˙ ( s ) ϕ ˙ ( s ) = 0 {:[(2.5)theta^(¨)(s)-(1)/(2)sin 2theta(s)phi^(˙)(s)phi^(˙)(s)=0],[(2.6)phi^(¨)(s)+2cot theta(s)theta^(˙)(s)phi^(˙)(s)=0]:}\begin{align*} \ddot{\theta}(s)-\frac{1}{2} \sin 2 \theta(s) \dot{\phi}(s) \dot{\phi}(s) & =0 \tag{2.5}\\ \ddot{\phi}(s)+2 \cot \theta(s) \dot{\theta}(s) \dot{\phi}(s) & =0 \tag{2.6} \end{align*}(2.5)θ¨(s)12sin2θ(s)ϕ˙(s)ϕ˙(s)=0(2.6)ϕ¨(s)+2cotθ(s)θ˙(s)ϕ˙(s)=0
Find the general solution to the geodesic equations.
b) Let M M MMM and Γ Γ Gamma\GammaΓ be as in a). Furthermore, let ( θ , ϕ ) = ( α s + β , ϕ 0 ) ( θ , ϕ ) = α s + β , ϕ 0 (theta,phi)=(alpha s+beta,phi_(0))(\theta, \phi)=\left(\alpha s+\beta, \phi_{0}\right)(θ,ϕ)=(αs+β,ϕ0), where s s sss is the curve parameter and α , β α , β alpha,beta\alpha, \betaα,β, and ϕ 0 ϕ 0 phi_(0)\phi_{0}ϕ0 are constants. Determine the parallel transport equations for a vector field X = ( X θ , X ϕ ) X = X θ , X ϕ X=(X^(theta),X^(phi))X=\left(X^{\theta}, X^{\phi}\right)X=(Xθ,Xϕ) and solve this set of equations. In addition, if u u uuu is the tangent vector ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) at the point ( θ , ϕ ) = ( π 4 , 0 ) ( θ , ϕ ) = π 4 , 0 (theta,phi)=((pi)/(4),0)(\theta, \phi)=\left(\frac{\pi}{4}, 0\right)(θ,ϕ)=(π4,0), then determine the parallel transported vector v v vvv at the point ( θ , ϕ ) = ( π 2 , 0 ) ( θ , ϕ ) = π 2 , 0 (theta,phi)=((pi)/(2),0)(\theta, \phi)=\left(\frac{\pi}{2}, 0\right)(θ,ϕ)=(π2,0).
Problem 2.10 Determine the shortest path on the conical surface r = a z r = a z r=-azr=-a zr=az which connects the points z = h , φ = 0 z = h , φ = 0 z=-h,varphi=0z=-h, \varphi=0z=h,φ=0 and z = h , φ = π / 2 z = h , φ = π / 2 z=-h,varphi=pi//2z=-h, \varphi=\pi / 2z=h,φ=π/2, where ( r , φ , z ) ( r , φ , z ) (r,varphi,z)(r, \varphi, z)(r,φ,z) are cylindrical coordinates and a > 0 a > 0 a > 0a>0a>0 and h > 0 h > 0 h > 0h>0h>0 constants.
Problem 2.11 A ship starts from a position in the Atlantic Ocean with coordinates 10 N 30 W 10 N 30 W 10^(@)N30^(@)W10^{\circ} \mathrm{N} 30^{\circ} \mathrm{W}10N30W (Cape Verde). It sails directly to the north to the 45 45 45^(@)45^{\circ}45 northern latitude (Azores, Portugal) and then it turns abruptly to the west and sails until it hits the 60 60 60^(@)60^{\circ}60 western longitude (Nova Scotia, Canada). Suppose a vector is parallel transported along the route of the ship (with the help of a gyroscope). Its initial direction is 45 45 45^(@)45^{\circ}45 (north-east). What is its final direction?
Problem 2.12 A vector is first parallel transported along a great circle on a sphere from a point A A AAA on the equator to the North Pole N N NNN, then again along a great circle from N N NNN to another point B B BBB on the equator, and finally, along the equator back to the point A A AAA. Use the standard Riemannian metric on the sphere and prove that the vector is rotated in the above process by an angle θ θ theta\thetaθ, which is directly proportional to the area of the geodesic triangle A N B A N B ANBA N BANB.
Problem 2.13 Let M = S 2 R 3 M = S 2 R 3 M=S^(2)subR^(3)M=\mathbb{S}^{2} \subset \mathbb{R}^{3}M=S2R3. Determine the metric g g ggg on M M MMM in terms of the spherical coordinates θ θ theta\thetaθ and ϕ ϕ phi\phiϕ. In particular, compute the inner product of the vectors ( 1 , 2 ) ( 1 , 2 ) (1,2)(1,2)(1,2) and ( 2 , 1 ) ( 2 , 1 ) (2,-1)(2,-1)(2,1) at the point ( θ , ϕ ) ( θ , ϕ ) (theta,phi)(\theta, \phi)(θ,ϕ).
Problem 2.14 Compute the Riemann curvature tensor R R RRR of the unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2.
Problem 2.15 Consider the vector fields
(2.7) X = x y y x and Y = x x + y y , (2.7) X = x y y x  and  Y = x x + y y , {:(2.7)X=x(del)/(del y)-y(del)/(del x)quad" and "quad Y=x(del)/(del x)+y(del)/(del y)",":}\begin{equation*} X=x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x} \quad \text { and } \quad Y=x \frac{\partial}{\partial x}+y \frac{\partial}{\partial y}, \tag{2.7} \end{equation*}(2.7)X=xyyx and Y=xx+yy,
in the x y x y xyx yxy-plane.
a) Determine the commutator [ X , Y ] [ X , Y ] [X,Y][X, Y][X,Y].
b) Assume that an affine connection in the plane satisfies X X = Y , Y Y = Y X X = Y , Y Y = Y grad_(X)X=-Y,grad_(Y)Y=Y\nabla_{X} X=-Y, \nabla_{Y} Y=YXX=Y,YY=Y, Y X = X Y X = X grad_(Y)X=X\nabla_{Y} X=XYX=X, and that the torsion tensor T T TTT vanishes. Compute the Riemann curvature tensor R R RRR.
Problem 2.16 Let x 1 x 1 x^(1)x^{1}x1 and x 2 x 2 x^(2)x^{2}x2 be a pair of local coordinates and
(2.8) X = x 2 x 1 x 1 x 2 , Y = x 1 x 1 + x 2 x 2 , (2.8) X = x 2 x 1 x 1 x 2 , Y = x 1 x 1 + x 2 x 2 , {:(2.8)X=x^(2)(del)/(delx^(1))-x^(1)(del)/(delx^(2))","quad Y=x^(1)(del)/(delx^(1))+x^(2)(del)/(delx^(2))",":}\begin{equation*} X=x^{2} \frac{\partial}{\partial x^{1}}-x^{1} \frac{\partial}{\partial x^{2}}, \quad Y=x^{1} \frac{\partial}{\partial x^{1}}+x^{2} \frac{\partial}{\partial x^{2}}, \tag{2.8} \end{equation*}(2.8)X=x2x1x1x2,Y=x1x1+x2x2,
be a pair of vector fields in R 2 { 0 } R 2 { 0 } R^(2)\\{0}\mathbb{R}^{2} \backslash\{0\}R2{0}. Assume that
X X = 0 , X Y = X + Y , (2.9) Y X = X Y , Y Y = 0 . X X = 0 , X Y = X + Y , (2.9) Y X = X Y , Y Y = 0 . {:[grad_(X)X=0","quadgrad_(X)Y=X+Y","],[(2.9)grad_(Y)X=X-Y","quadgrad_(Y)Y=0.]:}\begin{align*} & \nabla_{X} X=0, \quad \nabla_{X} Y=X+Y, \\ & \nabla_{Y} X=X-Y, \quad \nabla_{Y} Y=0 . \tag{2.9} \end{align*}XX=0,XY=X+Y,(2.9)YX=XY,YY=0.
Compute the components R 1 1 i j R 1 1 i j R^(1)_(1ij)R^{1}{ }_{1 i j}R11ij in the local coordinate basis, where i , j = 1 , 2 i , j = 1 , 2 i,j=1,2i, j=1,2i,j=1,2, of the Riemann curvature tensor.
Problem 2.17 A manifold M M M\mathcal{M}M of dimension 3 has a basis of orthonormal vector fields { L 1 , L 2 , L 3 } L 1 , L 2 , L 3 {L_(1),L_(2),L_(3)}\left\{L_{1}, L_{2}, L_{3}\right\}{L1,L2,L3} with commutation relations
(2.10) [ L i , L j ] = ϵ i j k L k , where i , j , k = 1 , 2 , 3 . (2.10) L i , L j = ϵ i j k L k ,  where  i , j , k = 1 , 2 , 3 . {:(2.10)[L_(i),L_(j)]=epsilon_(ijk)L_(k)","quad" where "i","j","k=1","2","3.:}\begin{equation*} \left[L_{i}, L_{j}\right]=\epsilon_{i j k} L_{k}, \quad \text { where } i, j, k=1,2,3 . \tag{2.10} \end{equation*}(2.10)[Li,Lj]=ϵijkLk, where i,j,k=1,2,3.
Determine the Levi-Civita connection i = L i ( 1 i 3 ) i = L i ( 1 i 3 ) grad_(i)=grad_(L_(i))(1 <= i <= 3)\nabla_{i}=\nabla_{L_{i}}(1 \leq i \leq 3)i=Li(1i3) and its Riemann curvature tensor R R RRR.
Hint: The Levi-Civita connection is the unique metric-compatible torsion-free connection. Use the symmetry properties of the Christoffel symbols coming from this, several times, to evaluate them.
Problem 2.18 Let x x xxx and y y yyy be local coordinates on a surface S S SSS with x + y 0 x + y 0 x+y!=0x+y \neq 0x+y0. Define a metric tensor g g ggg by g x x = 1 , g x y = g y x = x + y g x x = 1 , g x y = g y x = x + y g_(xx)=1,g_(xy)=g_(yx)=x+yg_{x x}=1, g_{x y}=g_{y x}=x+ygxx=1,gxy=gyx=x+y, and g y y = 1 + ( x + y ) 2 g y y = 1 + ( x + y ) 2 g_(yy)=1+(x+y)^(2)g_{y y}=1+(x+y)^{2}gyy=1+(x+y)2. Let grad\nabla be an affine connection defined by
(2.11) x x = ( x + y ) x y , (2.12) x y = [ 2 + ( x + y ) 2 ] x ( x + y ) y , (2.13) y x = ( x + y ) ( x + y + 1 ) x ( x + y + 1 ) y , (2.14) y y = { ( x + y + 1 ) [ 1 + ( x + y ) 2 ] + 1 } x ( x + y ) ( x + y + 1 ) y . (2.11) x x = ( x + y ) x y , (2.12) x y = 2 + ( x + y ) 2 x ( x + y ) y , (2.13) y x = ( x + y ) ( x + y + 1 ) x ( x + y + 1 ) y , (2.14) y y = ( x + y + 1 ) 1 + ( x + y ) 2 + 1 x ( x + y ) ( x + y + 1 ) y . {:[(2.11)grad_(x)del_(x)=(x+y)del_(x)-del_(y)","],[(2.12)grad_(x)del_(y)=[2+(x+y)^(2)]del_(x)-(x+y)del_(y)","],[(2.13)grad_(y)del_(x)=(x+y)(x+y+1)del_(x)-(x+y+1)del_(y)","],[(2.14)grad_(y)del_(y)={(x+y+1)[1+(x+y)^(2)]+1}del_(x)-(x+y)(x+y+1)del_(y).]:}\begin{align*} & \nabla_{x} \partial_{x}=(x+y) \partial_{x}-\partial_{y}, \tag{2.11}\\ & \nabla_{x} \partial_{y}=\left[2+(x+y)^{2}\right] \partial_{x}-(x+y) \partial_{y}, \tag{2.12}\\ & \nabla_{y} \partial_{x}=(x+y)(x+y+1) \partial_{x}-(x+y+1) \partial_{y}, \tag{2.13}\\ & \nabla_{y} \partial_{y}=\left\{(x+y+1)\left[1+(x+y)^{2}\right]+1\right\} \partial_{x}-(x+y)(x+y+1) \partial_{y} . \tag{2.14} \end{align*}(2.11)xx=(x+y)xy,(2.12)xy=[2+(x+y)2]x(x+y)y,(2.13)yx=(x+y)(x+y+1)x(x+y+1)y,(2.14)yy={(x+y+1)[1+(x+y)2]+1}x(x+y)(x+y+1)y.
a) Compute the Christoffel symbols in the orthonormal basis
(2.15) e 1 = x , e 2 = ( x + y ) x + y (2.15) e 1 = x , e 2 = ( x + y ) x + y {:(2.15)e_(1)=del_(x)","quade_(2)=-(x+y)del_(x)+del_(y):}\begin{equation*} e_{1}=\partial_{x}, \quad e_{2}=-(x+y) \partial_{x}+\partial_{y} \tag{2.15} \end{equation*}(2.15)e1=x,e2=(x+y)x+y
b) Consider the parallel transport of a pair of vectors starting from the point ( x , y ) = ( 1 , 1 ) ( x , y ) = ( 1 , 1 ) (x,y)=(1,1)(x, y)=(1,1)(x,y)=(1,1), counterclockwise along the full circle with center at ( x , y ) = ( 2 , 2 ) ( x , y ) = ( 2 , 2 ) (x,y)=(2,2)(x, y)=(2,2)(x,y)=(2,2) and radius r = 2 r = 2 r=sqrt2r=\sqrt{2}r=2. Assume that the initial angle between the vectors is π / 3 π / 3 pi//3\pi / 3π/3. What is the angle after the parallel transport around the loop?
Problem 2.19 Three ants are walking on a two-dimensional surface embedded in a flat three-dimensional Euclidean space as
(2.16) x = r cos ϕ , y = r sin ϕ , z = 2 3 r 3 / 2 (2.16) x = r cos ϕ , y = r sin ϕ , z = 2 3 r 3 / 2 {:(2.16)x=r cos phi","quad y=r sin phi","quad z=(2)/(3)r^(3//2):}\begin{equation*} x=r \cos \phi, \quad y=r \sin \phi, \quad z=\frac{2}{3} r^{3 / 2} \tag{2.16} \end{equation*}(2.16)x=rcosϕ,y=rsinϕ,z=23r3/2
Assume that the ants are walking on the surface along curves parametrized by λ λ lambda\lambdaλ such that
  • Ant #1: r = λ , ϕ = 0 r = λ , ϕ = 0 r=lambda,phi=0r=\lambda, \phi=0r=λ,ϕ=0;
  • Ant #2: r = λ 2 / 3 1 , ϕ = π / 2 , λ > 1 r = λ 2 / 3 1 , ϕ = π / 2 , λ > 1 r=lambda^(2//3)-1,phi=pi//2,quad lambda > 1r=\lambda^{2 / 3}-1, \phi=\pi / 2, \quad \lambda>1r=λ2/31,ϕ=π/2,λ>1;
  • Ant #3: r = λ 1 / 2 , ϕ = ln λ , λ > 0 r = λ 1 / 2 , ϕ = ln λ , λ > 0 r=lambda^(1//2),phi=ln lambda,quad lambda > 0r=\lambda^{1 / 2}, \phi=\ln \lambda, \quad \lambda>0r=λ1/2,ϕ=lnλ,λ>0.
    a) Compute the induced metric on the two-dimensional surface.
    b) Investigate if the ants are walking along geodesics or not.
Problem 2.20 Derive the explicit form of the geodesic equation on the hyperboloid x 2 + y 2 z 2 = a 2 x 2 + y 2 z 2 = a 2 x^(2)+y^(2)-z^(2)=a^(2)x^{2}+y^{2}-z^{2}=a^{2}x2+y2z2=a2 with x , y x , y x,yx, yx,y, and z z zzz being Cartesian coordinates on the flat Euclidean three-dimensional space (i.e., the metric is d s 2 = d x 2 + d y 2 + d z 2 d s 2 = d x 2 + d y 2 + d z 2 ds^(2)=dx^(2)+dy^(2)+dz^(2)d s^{2}=d x^{2}+d y^{2}+d z^{2}ds2=dx2+dy2+dz2 ) and a > 0 a > 0 a > 0a>0a>0 a constant. Using the coordinates r r rrr and φ φ varphi\varphiφ such that x = r cos ( φ ) x = r cos ( φ ) x=r cos(varphi)x=r \cos (\varphi)x=rcos(φ) and y = r sin ( φ ) y = r sin ( φ ) y=r sin(varphi)y=r \sin (\varphi)y=rsin(φ), compute also all Christoffel symbols for this hyperboloid.
Problem 2.21 Consider the surface ( c t ) 2 x 2 y 2 = K 2 ( c t ) 2 x 2 y 2 = K 2 (ct)^(2)-x^(2)-y^(2)=-K^(2)(c t)^{2}-x^{2}-y^{2}=-K^{2}(ct)2x2y2=K2 in the three-dimensional Minkowski space R 3 R 3 R^(3)\mathbb{R}^{3}R3 with metric signature +-- .
a) Compute the metric tensor on the surface in a suitable coordinate system. Is it positive definite? If not, of what type?
b) Find the geodesic equations. In particular, find a geodesic starting from the point ( 0 , 0 , K ) ( 0 , 0 , K ) (0,0,K)(0,0, K)(0,0,K) and going in the direction of the tangent vector ( c , 0 , 0 ) ( c , 0 , 0 ) (c,0,0)(c, 0,0)(c,0,0).
Problem 2.22 Consider the pseudo-Riemannian metric
(2.17) d s 2 = ( d x 1 ) 2 + ( d x 2 ) 2 ( d x 3 ) 2 ( d x 4 ) 2 (2.17) d s 2 = d x 1 2 + d x 2 2 d x 3 2 d x 4 2 {:(2.17)ds^(2)=(dx^(1))^(2)+(dx^(2))^(2)-(dx^(3))^(2)-(dx^(4))^(2):}\begin{equation*} d s^{2}=\left(d x^{1}\right)^{2}+\left(d x^{2}\right)^{2}-\left(d x^{3}\right)^{2}-\left(d x^{4}\right)^{2} \tag{2.17} \end{equation*}(2.17)ds2=(dx1)2+(dx2)2(dx3)2(dx4)2
in R 4 R 4 R^(4)\mathbb{R}^{4}R4. This induces a pseudo-Riemannian metric g g ggg on the surface
(2.18) S : ( x 1 ) 2 + ( x 2 ) 2 ( x 3 ) 2 ( x 4 ) 2 = 1 (2.18) S : x 1 2 + x 2 2 x 3 2 x 4 2 = 1 {:(2.18)S:quad(x^(1))^(2)+(x^(2))^(2)-(x^(3))^(2)-(x^(4))^(2)=1:}\begin{equation*} S: \quad\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}-\left(x^{4}\right)^{2}=1 \tag{2.18} \end{equation*}(2.18)S:(x1)2+(x2)2(x3)2(x4)2=1
a) Show that the metric g g ggg on S S SSS is Lorentzian, i.e., it has one timelike and two spacelike directions at each point.
b) Construct a pair of constants of motions for freely falling bodies by integrating the geodesic equations on S S SSS once.
Problem 2.23 The flow lines generated by a vector field X X XXX are smooth curves γ ( t ) γ ( t ) gamma(t)\gamma(t)γ(t) such that
(2.19) γ ˙ ( t ) = X ( γ ( t ) ) , (2.19) γ ˙ ( t ) = X ( γ ( t ) ) , {:(2.19)gamma^(˙)(t)=X(gamma(t))",":}\begin{equation*} \dot{\gamma}(t)=X(\gamma(t)), \tag{2.19} \end{equation*}(2.19)γ˙(t)=X(γ(t)),
along the curve. Assume that all flow lines for a vector field X X XXX are geodesics with respect to a connection determined by the Christoffel symbols Γ i j k Γ i j k Gamma_(ij)^(k)\Gamma_{i j}^{k}Γijk. Derive a set of partial differential equations for the components of X X XXX giving a necessary and sufficient condition for the above property of X X XXX.
Problem 2.24 Parametrize the points on the spin group SU ( 2 ) SU ( 2 ) SU(2)\mathrm{SU}(2)SU(2) as
(2.20) g ( x ) = e i ( x 1 σ 1 + x 2 σ 2 + x 3 σ 3 ) = 1 cos ( r ) + i sin ( r ) r x σ , (2.20) g ( x ) = e i x 1 σ 1 + x 2 σ 2 + x 3 σ 3 = 1 cos ( r ) + i sin ( r ) r x σ , {:(2.20)g(x)=e^(i(x^(1)sigma_(1)+x^(2)sigma_(2)+x^(3)sigma_(3)))=1cos(r)+i(sin(r))/(r)x*sigma",":}\begin{equation*} g(\boldsymbol{x})=e^{i\left(x^{1} \sigma_{1}+x^{2} \sigma_{2}+x^{3} \sigma_{3}\right)}=\mathbb{1} \cos (r)+i \frac{\sin (r)}{r} \boldsymbol{x} \cdot \boldsymbol{\sigma}, \tag{2.20} \end{equation*}(2.20)g(x)=ei(x1σ1+x2σ2+x3σ3)=1cos(r)+isin(r)rxσ,
where x = ( x 1 , x 2 , x 3 ) R 3 , r = | x | x = x 1 , x 2 , x 3 R 3 , r = | x | x=(x^(1),x^(2),x^(3))inR^(3),r=|x|\boldsymbol{x}=\left(x^{1}, x^{2}, x^{3}\right) \in \mathbb{R}^{3}, r=|\boldsymbol{x}|x=(x1,x2,x3)R3,r=|x| and the σ k σ k sigma_(k)\sigma_{k}σk 's are the Hermitian 2 × 2 2 × 2 2xx22 \times 22×2 Pauli matrices
(2.21) σ 1 = ( 0 1 1 0 ) , σ 2 = ( 0 i i 0 ) , and σ 3 = ( 1 0 0 1 ) . (2.21) σ 1 = 0 1 1 0 , σ 2 = 0 i i 0 ,  and  σ 3 = 1 0 0 1 . {:(2.21)sigma_(1)=([0,1],[1,0])","quadsigma_(2)=([0,-i],[i,0])","quad" and "quadsigma_(3)=([1,0],[0,-1]).:}\sigma_{1}=\left(\begin{array}{ll} 0 & 1 \tag{2.21}\\ 1 & 0 \end{array}\right), \quad \sigma_{2}=\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right), \quad \text { and } \quad \sigma_{3}=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) .(2.21)σ1=(0110),σ2=(0ii0), and σ3=(1001).
We can identify the point g ( x ) g ( x ) g(x)g(x)g(x) as a point on the 3-dimensional unit sphere S 3 R 4 S 3 R 4 S^(3)subR^(4)\mathbb{S}^{3} \subset \mathbb{R}^{4}S3R4, the first coordinate is cos ( r ) cos ( r ) cos(r)\cos (r)cos(r) and the remaining three coordinates are sin ( r ) x / r sin ( r ) x / r sin(r)x//r\sin (r) \boldsymbol{x} / rsin(r)x/r. Show that the 1-parameter subgroups t e ita σ t e ita  σ t|->e^("ita "*sigma)t \mapsto e^{\text {ita } \cdot \sigma}teita σ, where a R 3 a R 3 a inR^(3)\boldsymbol{a} \in \mathbb{R}^{3}aR3, are geodesics with respect to the standard metric on S 3 S 3 S^(3)\mathbb{S}^{3}S3 coming from the Euclidean metric in R 4 R 4 R^(4)\mathbb{R}^{4}R4.
Hint: It is more convenient to use the Euler-Lagrange equations coming from the metric element (derive the formula!) in terms of the angular coordinates θ , ϕ θ , ϕ theta,phi\theta, \phiθ,ϕ of the vector x R 3 x R 3 x inR^(3)\boldsymbol{x} \in \mathbb{R}^{3}xR3 and the radial coordinate r r rrr.
Problem 2.25 a) Derive the relation between the Christoffel symbols Γ μ ν λ Γ μ ν λ Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}Γμνλ and the metric tensor g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν from the following conditions: (i) D λ g μ ν = 0 D λ g μ ν = 0 D_(lambda)g_(mu nu)=0D_{\lambda} g_{\mu \nu}=0Dλgμν=0, where D λ D λ D_(lambda)D_{\lambda}Dλ is the covariant derivative, and (ii) Γ μ ν λ = Γ ν μ λ Γ μ ν λ = Γ ν μ λ Gamma_(mu nu)^(lambda)=Gamma_(nu mu)^(lambda)\Gamma_{\mu \nu}^{\lambda}=\Gamma_{\nu \mu}^{\lambda}Γμνλ=Γνμλ. (The result is the so-called "fundamental theorem" in Riemannian geometry.)
b) Consider the vector field ( V μ ) = ( x , t ) V μ = ( x , t ) (V^(mu))=(x,-t)\left(V^{\mu}\right)=(x,-t)(Vμ)=(x,t), i.e., V 0 = x V 0 = x V^(0)=xV^{0}=xV0=x and V 1 = t V 1 = t V^(1)=-tV^{1}=-tV1=t, in twodimensional Minkowski spacetime with coordinates ( x μ ) = ( t , x ) x μ = ( t , x ) (x^(mu))=(t,x)\left(x^{\mu}\right)=(t, x)(xμ)=(t,x) and metric d s 2 = d s 2 = ds^(2)=d s^{2}=ds2= d t 2 d x 2 d t 2 d x 2 dt^(2)-dx^(2)d t^{2}-d x^{2}dt2dx2. Compute all components of the tensor T μ ν = D μ V v T μ ν = D μ V v T_(mu)^(nu)=D_(mu)V^(v)T_{\mu}{ }^{\nu}=D_{\mu} V^{v}Tμν=DμVv in this coordinate system. Compute also the component T 0 1 T 0 1 T_(0)^(1)T_{0}{ }^{1}T01 of this tensor in Rindler coordinates ( x μ ) = ( λ , a ) x μ = ( λ , a ) (x^('mu))=(lambda,a)\left(x^{\prime \mu}\right)=(\lambda, a)(xμ)=(λ,a) defined as
(2.22) t = a sinh ( λ ) , x = a cosh ( λ ) (2.22) t = a sinh ( λ ) , x = a cosh ( λ ) {:(2.22)t=a sinh(lambda)","quad x=a cosh(lambda):}\begin{equation*} t=a \sinh (\lambda), \quad x=a \cosh (\lambda) \tag{2.22} \end{equation*}(2.22)t=asinh(λ),x=acosh(λ)
Problem 2.26 a) Write the transformation law for a tensor with components S μ ν S μ ν S^(mu nu)S^{\mu \nu}Sμν under a general coordinate transformation x μ x μ x μ x μ x^(mu)|->x^('mu)x^{\mu} \mapsto x^{\prime \mu}xμxμ, i.e., give a general formula for S μ ν S μ ν S^('mu nu)S^{\prime \mu \nu}Sμν.
b) Write D μ S μ v D μ S μ v D_(mu)S^(mu v)D_{\mu} S^{\mu v}DμSμv in terms of partial derivatives and Christoffel symbols.
c) Consider the tensor with components S 12 = S 21 = 2 x y S 12 = S 21 = 2 x y S^(12)=-S^(21)=2xyS^{12}=-S^{21}=2 x yS12=S21=2xy and S 11 = S 22 = 0 S 11 = S 22 = 0 S^(11)=S^(22)=0S^{11}=S^{22}=0S11=S22=0 on the two-dimensional plane with coordinates ( x μ ) = ( x , y ) x μ = ( x , y ) (x^(mu))=(x,y)\left(x^{\mu}\right)=(x, y)(xμ)=(x,y) and metric d s 2 = d s 2 = ds^(2)=d s^{2}=ds2= d x 2 + d y 2 d x 2 + d y 2 dx^(2)+dy^(2)d x^{2}+d y^{2}dx2+dy2. (i) Compute the components S μ ν S μ ν S^('mu nu)S^{\prime \mu \nu}Sμν of this tensor in polar coordinates ( x μ ) = ( r , φ ) x μ = ( r , φ ) (x^('mu))=(r,varphi)\left(x^{\prime \mu}\right)=(r, \varphi)(xμ)=(r,φ). (ii) Compute D μ S μ ν D μ S μ ν D_(mu)S^(mu nu)D_{\mu} S^{\mu \nu}DμSμν.
Problem 2.27 Consider AdS 2 AdS 2 AdS_(2)\mathrm{AdS}_{2}AdS2 which is a two-dimensional curved spacetime with coordinates ( x μ ) = ( x 0 , x 1 ) = ( t , r ) x μ = x 0 , x 1 = ( t , r ) (x^(mu))=(x^(0),x^(1))=(t,r)\left(x^{\mu}\right)=\left(x^{0}, x^{1}\right)=(t, r)(xμ)=(x0,x1)=(t,r) and the metric given by
(2.23) d s 2 = v [ ( r 2 1 ) d t 2 1 r 2 1 d r 2 ] , (2.23) d s 2 = v r 2 1 d t 2 1 r 2 1 d r 2 , {:(2.23)ds^(2)=v[(r^(2)-1)dt^(2)-(1)/(r^(2)-1)dr^(2)]",":}\begin{equation*} d s^{2}=v\left[\left(r^{2}-1\right) d t^{2}-\frac{1}{r^{2}-1} d r^{2}\right], \tag{2.23} \end{equation*}(2.23)ds2=v[(r21)dt21r21dr2],
where v > 0 v > 0 v > 0v>0v>0 is some constant. Consider also on AdS 2 AdS 2 AdS_(2)\mathrm{AdS}_{2}AdS2 and in the coordinates x μ x μ x^(mu)x^{\mu}xμ, a tensor field S μ ν S μ ν S_(mu nu)S_{\mu \nu}Sμν with the following components S 00 = a ( r 2 1 ) , S 11 = a / ( r 2 1 ) S 00 = a r 2 1 , S 11 = a / r 2 1 S_(00)=a(r^(2)-1),S_(11)=-a//(r^(2)-1)S_{00}=a\left(r^{2}-1\right), S_{11}=-a /\left(r^{2}-1\right)S00=a(r21),S11=a/(r21), and S 01 = S 10 = 0 S 01 = S 10 = 0 S_(01)=S_(10)=0S_{01}=S_{10}=0S01=S10=0 for some constant a > 0 a > 0 a > 0a>0a>0.
a) Compute all Christoffel symbols for AdS 2 AdS 2 AdS_(2)\mathrm{AdS}_{2}AdS2 in the coordinates x μ x μ x^(mu)x^{\mu}xμ.
b) Compute S μ ν S μ ν S^(mu nu)S^{\mu \nu}Sμν and D μ S μ ν D μ S μ ν D_(mu)S^(mu nu)D_{\mu} S^{\mu \nu}DμSμν.
c) Consider another coordinate system ( x μ ) = ( x 0 , x 1 ) = ( θ , η ) x μ = x 0 , x 1 = ( θ , η ) (x^('mu))=(x^('0),x^('1))=(theta,eta)\left(x^{\prime \mu}\right)=\left(x^{\prime 0}, x^{\prime 1}\right)=(\theta, \eta)(xμ)=(x0,x1)=(θ,η) on AdS 2 AdS 2 AdS_(2)\operatorname{AdS}_{2}AdS2 defined by θ = a t θ = a t theta=at\theta=a tθ=at and r = cosh ( η ) r = cosh ( η ) r=cosh(eta)r=\cosh (\eta)r=cosh(η) with a a aaa being the same constant as above. Transform the tensor S μ ν S μ ν S_(mu nu)S_{\mu \nu}Sμν to this new coordinate system, i.e., compute S μ ν S μ ν S_(mu nu)^(')S_{\mu \nu}^{\prime}Sμν.
Problem 2.28 Consider the so-called Rindler coordinate system ( x μ ) = x μ = (x^('mu))=\left(x^{\prime \mu}\right)=(xμ)= ( x 0 , x 1 ) = ( λ , a ) x 0 , x 1 = ( λ , a ) (x^('0),x^('1))=(lambda,a)\left(x^{\prime 0}, x^{\prime 1}\right)=(\lambda, a)(x0,x1)=(λ,a) in two-dimensional Minkowski space defined by
(2.24) t = a sinh ( λ ) , x = a cosh ( λ ) , a > 0 , λ R (2.24) t = a sinh ( λ ) , x = a cosh ( λ ) , a > 0 , λ R {:(2.24)t=a sinh(lambda)","quad x=a cosh(lambda)","quad a > 0","lambda inR:}\begin{equation*} t=a \sinh (\lambda), \quad x=a \cosh (\lambda), \quad a>0, \lambda \in \mathbb{R} \tag{2.24} \end{equation*}(2.24)t=asinh(λ),x=acosh(λ),a>0,λR
where ( x μ ) = ( t , x ) x μ = ( t , x ) (x^(mu))=(t,x)\left(x^{\mu}\right)=(t, x)(xμ)=(t,x) are the usual coordinates, i.e., d s 2 = d t 2 d x 2 d s 2 = d t 2 d x 2 ds^(2)=dt^(2)-dx^(2)d s^{2}=d t^{2}-d x^{2}ds2=dt2dx2.
a) Find the metric tensor and the Christoffel symbols in this coordinate system.
b) Determine expressions for the divergence of a vector field and the Laplacian of a scalar field in two-dimensional Minkowski space in Rindler coordinates.
c) A tensor of rank two on two-dimensional Minkowski space has the following components in the coordinates ( x μ ) = ( t , x ) : T 0 0 = T 1 1 = x 2 t 2 x μ = ( t , x ) : T 0 0 = T 1 1 = x 2 t 2 (x^(mu))=(t,x):T_(0)^(0)=-T_(1)^(1)=x^(2)-t^(2)\left(x^{\mu}\right)=(t, x): T_{0}^{0}=-T_{1}^{1}=x^{2}-t^{2}(xμ)=(t,x):T00=T11=x2t2 and T 0 1 = T 0 1 = T_(0)^(1)=T_{0}^{1}=T01= T 1 0 = 0 T 1 0 = 0 T_(1)^(0)=0T_{1}^{0}=0T10=0. Compute the component T 0 0 T 0 0 T^('0)_(0)T^{\prime 0}{ }_{0}T00, i.e., T μ v T μ v T^('mu)_(v)T^{\prime \mu}{ }_{v}Tμv for μ = v = 0 μ = v = 0 mu=v=0\mu=v=0μ=v=0, of this tensor in Rindler coordinates.
Problem 2.29 Consider the metric
(2.25) d s 2 = d t 2 d r 2 r 2 d ϕ 2 (2.25) d s 2 = d t 2 d r 2 r 2 d ϕ 2 {:(2.25)ds^(2)=dt^(2)-dr^(2)-r^(2)dphi^(2):}\begin{equation*} d s^{2}=d t^{2}-d r^{2}-r^{2} d \phi^{2} \tag{2.25} \end{equation*}(2.25)ds2=dt2dr2r2dϕ2
Find expressions for the covariant derivative μ V v μ V v grad_(mu)V_(v)\nabla_{\mu} V_{v}μVv and the divergence μ V μ μ V μ grad_(mu)V^(mu)\nabla_{\mu} V^{\mu}μVμ.
Problem 2.30 The parallel transport of a vector V μ V μ V^(mu)V^{\mu}Vμ along a curve parametrized by λ λ lambda\lambdaλ is given by the condition
(2.26) d x μ d λ μ V ν = 0 . (2.26) d x μ d λ μ V ν = 0 . {:(2.26)(dx^(mu))/(d lambda)*grad_(mu)V^(nu)=0.:}\begin{equation*} \frac{d x^{\mu}}{d \lambda} \cdot \nabla_{\mu} V^{\nu}=0 . \tag{2.26} \end{equation*}(2.26)dxμdλμVν=0.
a) Obtain the geodesic equation
(2.27) x ¨ μ + Γ ν λ μ x ˙ ν x ˙ λ = 0 (2.27) x ¨ μ + Γ ν λ μ x ˙ ν x ˙ λ = 0 {:(2.27)x^(¨)^(mu)+Gamma_(nu lambda)^(mu)x^(˙)^(nu)x^(˙)^(lambda)=0:}\begin{equation*} \ddot{x}^{\mu}+\Gamma_{\nu \lambda}^{\mu} \dot{x}^{\nu} \dot{x}^{\lambda}=0 \tag{2.27} \end{equation*}(2.27)x¨μ+Γνλμx˙νx˙λ=0
from parallel transporting a tangent vector.
b) If one allows the tangent vector T μ T μ T^(mu)T^{\mu}Tμ to change in size, one can generalize the condition d x μ d λ μ V ν = 0 d x μ d λ μ V ν = 0 (dx^(mu))/(d lambda)grad_(mu)V^(nu)=0\frac{d x^{\mu}}{d \lambda} \nabla_{\mu} V^{\nu}=0dxμdλμVν=0 to
(2.28) T μ μ T v = α T v , where T μ = d x μ d λ . (2.28) T μ μ T v = α T v ,  where  T μ = d x μ d λ . {:(2.28)T^(mu)grad_(mu)T^(v)=alphaT^(v)","quad" where "T^(mu)=(dx^(mu))/(d lambda).:}\begin{equation*} T^{\mu} \nabla_{\mu} T^{v}=\alpha T^{v}, \quad \text { where } T^{\mu}=\frac{d x^{\mu}}{d \lambda} . \tag{2.28} \end{equation*}(2.28)TμμTv=αTv, where Tμ=dxμdλ.
Show that one can get back the original condition, where one has zero on the righthand side, by making a reparametrization λ τ ( λ ) λ τ ( λ ) lambda rarr tau(lambda)\lambda \rightarrow \tau(\lambda)λτ(λ). Show also how the geodesic equation is modified by the extra term on the right-hand side.
Problem 2.31 A conformal transformation of the metric tensor is defined as
(2.29) g μ ν f ( x ) g μ ν (2.29) g μ ν f ( x ) g μ ν {:(2.29)g_(mu nu)longrightarrow f(x)g_(mu nu):}\begin{equation*} g_{\mu \nu} \longrightarrow f(x) g_{\mu \nu} \tag{2.29} \end{equation*}(2.29)gμνf(x)gμν
for an arbitrary positive function f f fff.
a) Show that a conformal transformation preserves the angle between any two vectors.
b) A null curve is a curve for which all tangent vectors are null vectors. Show that all null curves remain null curves after performing a conformal transformation.
Problem 2.32 A sphere is described locally by the two coordinates θ θ theta\thetaθ and φ φ varphi\varphiφ is embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3 according to
(2.30) x = R cos ( φ ) sin ( θ ) , y = R sin ( φ ) sin ( θ ) , z = R α cos ( θ ) , (2.30) x = R cos ( φ ) sin ( θ ) , y = R sin ( φ ) sin ( θ ) , z = R α cos ( θ ) , {:(2.30)x=R cos(varphi)sin(theta)","quad y=R sin(varphi)sin(theta)","quad z=R alpha cos(theta)",":}\begin{equation*} x=R \cos (\varphi) \sin (\theta), \quad y=R \sin (\varphi) \sin (\theta), \quad z=R \alpha \cos (\theta), \tag{2.30} \end{equation*}(2.30)x=Rcos(φ)sin(θ),y=Rsin(φ)sin(θ),z=Rαcos(θ),
where R R RRR and 0 < α < 1 0 < α < 1 0 < alpha < 10<\alpha<10<α<1 are constants (note that while the topology of this manifold is a sphere, this is not the standard embedding of the sphere in Euclidean space). Compute the induced metric tensor and the Christoffel symbols.

2.2 Christoffel Symbols, Riemann and Ricci Tensors, and Einstein's Equations

Problem 2.33 a) Let Γ μ ν λ Γ μ ν λ Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}Γμνλ be the Levi-Civita connection associated to a metric tensor g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν. Show that Γ μ ν μ = 1 2 g ¯ 1 ν g ¯ Γ μ ν μ = 1 2 g ¯ 1 ν g ¯ Gamma_(mu nu)^(mu)=(1)/(2) bar(g)^(-1)del_(nu) bar(g)\Gamma_{\mu \nu}^{\mu}=\frac{1}{2} \bar{g}^{-1} \partial_{\nu} \bar{g}Γμνμ=12g¯1νg¯, where g ¯ = det ( g μ ν ) g ¯ = det g μ ν bar(g)=det(g_(mu nu))\bar{g}=\operatorname{det}\left(g_{\mu \nu}\right)g¯=det(gμν).
b) Derive from the definition of covariant differentiation the transformation rule for the Christoffel symbols with respect to general coordinate transformations.
c) Show directly from the definition of parallel transport that in a parallel transport defined by the Levi-Civita connection, Γ μ ν λ = 1 2 g λ ω ( μ g ν ω + ν g μ ω ω g μ ν ) Γ μ ν λ = 1 2 g λ ω μ g ν ω + ν g μ ω ω g μ ν Gamma_(mu nu)^(lambda)=(1)/(2)g^(lambda omega)(del_(mu)g_(nu omega)+del_(nu)g_(mu omega)-del_(omega)g_(mu nu))\Gamma_{\mu \nu}^{\lambda}=\frac{1}{2} g^{\lambda \omega}\left(\partial_{\mu} g_{\nu \omega}+\partial_{\nu} g_{\mu \omega}-\partial_{\omega} g_{\mu \nu}\right)Γμνλ=12gλω(μgνω+νgμωωgμν), the length of a parallel transported vector is constant.
Problem 2.34 For any two vector fields U μ U μ U^(mu)U^{\mu}Uμ and V μ V μ V^(mu)V^{\mu}Vμ on a manifold with metric g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν equipped with the Levi-Civita connection, show that if
(2.31) W ν = U μ μ V v , (2.31) W ν = U μ μ V v , {:(2.31)W^(nu)=U^(mu)grad_(mu)V^(v)",":}\begin{equation*} W^{\nu}=U^{\mu} \nabla_{\mu} V^{v}, \tag{2.31} \end{equation*}(2.31)Wν=UμμVv,
then
(2.32) W v g v σ W σ = U μ μ V v (2.32) W v g v σ W σ = U μ μ V v {:(2.32)W_(v)-=g_(v sigma)W^(sigma)=U^(mu)grad_(mu)V_(v):}\begin{equation*} W_{v} \equiv g_{v \sigma} W^{\sigma}=U^{\mu} \nabla_{\mu} V_{v} \tag{2.32} \end{equation*}(2.32)WvgvσWσ=UμμVv
Problem 2.35 The nonzero Christoffel symbols on a unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2 in spherical coordinates are
(2.33) Γ ϕ ϕ θ = 1 2 sin 2 θ , Γ θ ϕ ϕ = Γ ϕ θ ϕ = cot θ . (2.33) Γ ϕ ϕ θ = 1 2 sin 2 θ , Γ θ ϕ ϕ = Γ ϕ θ ϕ = cot θ . {:(2.33)Gamma_(phi phi)^(theta)=-(1)/(2)sin 2theta","quadGamma_(theta phi)^(phi)=Gamma_(phi theta)^(phi)=cot theta.:}\begin{equation*} \Gamma_{\phi \phi}^{\theta}=-\frac{1}{2} \sin 2 \theta, \quad \Gamma_{\theta \phi}^{\phi}=\Gamma_{\phi \theta}^{\phi}=\cot \theta . \tag{2.33} \end{equation*}(2.33)Γϕϕθ=12sin2θ,Γθϕϕ=Γϕθϕ=cotθ.
a) Compute the Christoffel symbols Γ θ i j Γ θ i j Gamma_(theta i)^(j)\Gamma_{\theta i}^{j}Γθij and Γ ϕ i j Γ ϕ i j Gamma_(phi i)^(j)\Gamma_{\phi i}^{j}Γϕij in the orthonormal basis e 1 = θ e 1 = θ e_(1)=del_(theta)e_{1}=\partial_{\theta}e1=θ, e 2 = 1 sin θ ϕ , θ e i = Γ θ i j e j e 2 = 1 sin θ ϕ , θ e i = Γ θ i j e j e_(2)=(1)/(sin theta)del_(phi),grad_(theta)e_(i)=Gamma_(theta i)^(j)e_(j)e_{2}=\frac{1}{\sin \theta} \partial_{\phi}, \nabla_{\theta} e_{i}=\Gamma_{\theta i}^{j} e_{j}e2=1sinθϕ,θei=Γθijej, and ϕ e i = Γ ϕ i j e j ϕ e i = Γ ϕ i j e j grad_(phi)e_(i)=Gamma_(phi i)^(j)e_(j)\nabla_{\phi} e_{i}=\Gamma_{\phi i}^{j} e_{j}ϕei=Γϕijej.
b) Prove that the parallel transport of a vector u = u 1 e 1 + u 2 e 2 = ( u 1 u 2 ) u = u 1 e 1 + u 2 e 2 = ( u 1 u 2 ) u=u_(1)e_(1)+u_(2)e_(2)=((u_(1))/(u_(2)))u=u_{1} e_{1}+u_{2} e_{2}=\binom{u_{1}}{u_{2}}u=u1e1+u2e2=(u1u2) around a closed loop γ ( t ) γ ( t ) gamma(t)\gamma(t)γ(t) on S 2 S 2 S^(2)\mathbb{S}^{2}S2 is given by the operation u = R u u = R u u^(')=Ruu^{\prime}=R uu=Ru, where R R RRR is a rotation by an angle Ω Ω Omega\OmegaΩ equal to the area of the region bounded by the loop γ γ gamma\gammaγ.
Hint: First, write the solution as a line integral of the Christoffel symbols around the loop, and then, apply Green's formula in the plane.
Problem 2.36 The Christoffel symbols for the flat Euclidean metric in R 3 R 3 R^(3)\mathbb{R}^{3}R3 vanish. Compute the Christoffel symbols in the spherical coordinates ( r , θ , φ ) ( r , θ , φ ) (r,theta,varphi)(r, \theta, \varphi)(r,θ,φ).
Problem 2.37 A sphere can be projected onto a plane using stereographic projection. We use the metric
(2.34) d s 2 = R 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (2.34) d s 2 = R 2 d θ 2 + sin 2 θ d ϕ 2 {:(2.34)ds^(2)=R^(2)(dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*} d s^{2}=R^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{2.34} \end{equation*}(2.34)ds2=R2(dθ2+sin2θdϕ2)
for the sphere. The x x xxx and y y yyy coordinates of the plane can be expressed in terms of the spherical coordinates as
(2.35) x = 2 R tan ( θ / 2 ) cos ϕ , y = 2 R tan ( θ / 2 ) sin ϕ (2.35) x = 2 R tan ( θ / 2 ) cos ϕ , y = 2 R tan ( θ / 2 ) sin ϕ {:(2.35)x=2R tan(theta//2)cos phi","quad y=2R tan(theta//2)sin phi:}\begin{equation*} x=2 R \tan (\theta / 2) \cos \phi, \quad y=2 R \tan (\theta / 2) \sin \phi \tag{2.35} \end{equation*}(2.35)x=2Rtan(θ/2)cosϕ,y=2Rtan(θ/2)sinϕ
a) Express the metric of the sphere in terms of the x x xxx and y y yyy coordinates.
b) Compute the Christoffel symbols using the metric obtained in a).
c) Draw a picture illustrating the stereographic projection.
Problem 2.38 A two-dimensional hyperbolic subspace x 2 + y 2 t 2 = 1 , z = 0 x 2 + y 2 t 2 = 1 , z = 0 x^(2)+y^(2)-t^(2)=1,z=0x^{2}+y^{2}-t^{2}=1, z=0x2+y2t2=1,z=0 is embedded into the four-dimensional Minkowski space.
a) Parametrize the surface using only two parameters.
b) Compute the induced metric on the subspace.
c) Compute the Christoffel symbols in the subspace.
Problem 2.39 a) Starting from the definition of the curvature tensor
(2.36) R ( X , Y ) Z = [ X , Y ] Z [ X , Y ] Z (2.36) R ( X , Y ) Z = X , Y Z [ X , Y ] Z {:(2.36)R(X","Y)Z=[grad_(X),grad_(Y)]Z-grad_([X,Y])Z:}\begin{equation*} R(X, Y) Z=\left[\nabla_{X}, \nabla_{Y}\right] Z-\nabla_{[X, Y]} Z \tag{2.36} \end{equation*}(2.36)R(X,Y)Z=[X,Y]Z[X,Y]Z
derive the formula for the components R ω μ ν λ R ω μ ν λ R^(omega)_(mu nu lambda)R^{\omega}{ }_{\mu \nu \lambda}Rωμνλ in terms of the Christoffel symbols. Prove the first Bianchi identity
(2.37) R ω μ ν λ + R ν λ μ ω + R λ μ ν ω = 0 , (2.37) R ω μ ν λ + R ν λ μ ω + R λ μ ν ω = 0 , {:(2.37)R^(omega)_(mu nu lambda)+R_(nu lambda mu)^(omega)+R_(lambda mu nu)^(omega)=0",":}\begin{equation*} R^{\omega}{ }_{\mu \nu \lambda}+R_{\nu \lambda \mu}^{\omega}+R_{\lambda \mu \nu}^{\omega}=0, \tag{2.37} \end{equation*}(2.37)Rωμνλ+Rνλμω+Rλμνω=0,
in the case when the torsion T = 0 T = 0 T=0T=0T=0.
b) Prove the second Bianchi identity
(2.38) R α β μ ν ; λ + R α β ν λ ; μ + R α β λ μ ; ν = 0 (2.38) R α β μ ν ; λ + R α β ν λ ; μ + R α β λ μ ; ν = 0 {:(2.38)R_(alpha beta mu nu;lambda)+R_(alpha beta nu lambda;mu)+R_(alpha beta lambda mu;nu)=0:}\begin{equation*} R_{\alpha \beta \mu \nu ; \lambda}+R_{\alpha \beta \nu \lambda ; \mu}+R_{\alpha \beta \lambda \mu ; \nu}=0 \tag{2.38} \end{equation*}(2.38)Rαβμν;λ+Rαβνλ;μ+Rαβλμ;ν=0
and use this to show that the covariant derivative of the energy-momentum tensor T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν in Einstein's equations
(2.39) G μ ν = 8 π G c 4 T μ ν (2.39) G μ ν = 8 π G c 4 T μ ν {:(2.39)G^(mu nu)=(8pi G)/(c^(4))T^(mu nu):}\begin{equation*} G^{\mu \nu}=\frac{8 \pi G}{c^{4}} T^{\mu \nu} \tag{2.39} \end{equation*}(2.39)Gμν=8πGc4Tμν
vanishes.
Motivate that the vanishing of the covariant derivative of T μ v T μ v T^(mu v)T^{\mu v}Tμv coincides with local energy-momentum conservation for flat spacetime.
Problem 2.40 Derive the formula relating the Riemann curvature tensor to the parallel transport around an infinitesimal parallelogram.
Problem 2.41 Fix a metric on the paraboloid z = x 2 + y 2 z = x 2 + y 2 z=x^(2)+y^(2)z=x^{2}+y^{2}z=x2+y2 induced by the standard Euclidean metric in R 3 R 3 R^(3)\mathbb{R}^{3}R3. Compute the components of the Riemann curvature tensor on the paraboloid.
Hint: Use polar coordinates in the x y x y xyx yxy-plane.
Problem 2.42 Consider the two-dimensional metric
(2.40) d s 2 = r 2 ( d r 2 + r 2 d ϕ 2 ) (2.40) d s 2 = r 2 d r 2 + r 2 d ϕ 2 {:(2.40)ds^(2)=r^(2)(dr^(2)+r^(2)dphi^(2)):}\begin{equation*} d s^{2}=r^{2}\left(d r^{2}+r^{2} d \phi^{2}\right) \tag{2.40} \end{equation*}(2.40)ds2=r2(dr2+r2dϕ2)
a) Calculate the component R r ϕ r ϕ R r ϕ r ϕ R^(r)_(phi r phi)R^{r}{ }_{\phi r \phi}Rrϕrϕ of the Riemann tensor.
b) In flat Euclidean space, the relation between area and circumference of a circle is C 2 = 4 π A C 2 = 4 π A C^(2)=4pi AC^{2}=4 \pi AC2=4πA. What is the relation for a circle around the origin for the above metric? The area is given by the integral det ( g ) d r d ϕ det ( g ) d r d ϕ intsqrt(det(g))drd phi\int \sqrt{\operatorname{det}(g)} d r d \phidet(g)drdϕ.
Problem 2.43 Let M M MMM be a Lorentzian manifold of dimension n = 3 n = 3 n=3n=3n=3. Assume that there is an orthogonal basis of vector fields X , Y , Z X , Y , Z X,Y,ZX, Y, ZX,Y,Z such that
  1. g ( X , X ) = g ( Y , Y ) = g ( Z , Z ) = 1 g ( X , X ) = g ( Y , Y ) = g ( Z , Z ) = 1 g(X,X)=g(Y,Y)=-g(Z,Z)=-1g(X, X)=g(Y, Y)=-g(Z, Z)=-1g(X,X)=g(Y,Y)=g(Z,Z)=1,
  2. [ X , Y ] = Z , [ Y , Z ] = X , [ Z , X ] = Y [ X , Y ] = Z , [ Y , Z ] = X , [ Z , X ] = Y [X,Y]=-Z,[Y,Z]=X,[Z,X]=Y[X, Y]=-Z,[Y, Z]=X,[Z, X]=Y[X,Y]=Z,[Y,Z]=X,[Z,X]=Y,
    where g g ggg is the metric tensor. Compute the Christoffel symbols of the Levi-Civita connection and the Riemann curvature tensor in this basis.
    Hint: Use the symmetry properties of the Christoffel symbols coming from the torsion-free property of the connection together with X g = Y g = Z g = 0 X g = Y g = Z g = 0 grad_(X)g=grad_(Y)g=grad_(Z)g=0\nabla_{X} g=\nabla_{Y} g=\nabla_{Z} g=0Xg=Yg=Zg=0.
Problem 2.44 a) Show that the Ricci tensor R μ ν = R λ μ λ ν R μ ν = R λ μ λ ν R_(mu nu)=R^(lambda)_(mu lambda nu)R_{\mu \nu}=R^{\lambda}{ }_{\mu \lambda \nu}Rμν=Rλμλν (and thus also the Einstein tensor) is symmetric when the Riemann curvature tensor R α μ β v R α μ β v R^(alpha)_(mu beta v)R^{\alpha}{ }_{\mu \beta v}Rαμβv has been constructed from a metric.
b) Show that in two spacetime dimensions the tensor R μ ν k g μ ν R R μ ν k g μ ν R R_(mu nu)-kg_(mu nu)RR_{\mu \nu}-k g_{\mu \nu} RRμνkgμνR vanishes for some number k k kkk. Determine k k kkk.
c) Show that any metric in a 1+1-dimensional spacetime satisfies Einstein's equations in vacuum ( T μ ν = 0 T μ ν = 0 T_(mu nu)=0T_{\mu \nu}=0Tμν=0 ), i.e, G μ ν = 0 G μ ν = 0 G_(mu nu)=0G_{\mu \nu}=0Gμν=0.
Hint: Use the (anti)symmetries
(2.41) R α β μ ν = R β α μ ν = R α β v μ = R μ ν α β , (2.41) R α β μ ν = R β α μ ν = R α β v μ = R μ ν α β , {:(2.41)R_(alpha beta mu nu)=-R_(beta alpha mu nu)=-R_(alpha beta v mu)=R_(mu nu alpha beta)",":}\begin{equation*} R_{\alpha \beta \mu \nu}=-R_{\beta \alpha \mu \nu}=-R_{\alpha \beta v \mu}=R_{\mu \nu \alpha \beta}, \tag{2.41} \end{equation*}(2.41)Rαβμν=Rβαμν=Rαβvμ=Rμναβ,
of the Riemann curvature tensor.
Problem 2.45 Consider the two-dimensional curved spacetime with the metric given by
(2.42) d s 2 = 1 y 2 ( d t 2 d y 2 ) (2.42) d s 2 = 1 y 2 d t 2 d y 2 {:(2.42)ds^(2)=(1)/(y^(2))(dt^(2)-dy^(2)):}\begin{equation*} d s^{2}=\frac{1}{y^{2}}\left(d t^{2}-d y^{2}\right) \tag{2.42} \end{equation*}(2.42)ds2=1y2(dt2dy2)
in coordinates ( x μ ) = ( x 0 , x 1 ) = ( t , y ) x μ = x 0 , x 1 = ( t , y ) (x^(mu))=(x^(0),x^(1))=(t,y)\left(x^{\mu}\right)=\left(x^{0}, x^{1}\right)=(t, y)(xμ)=(x0,x1)=(t,y) with t R t R t inRt \in \mathbb{R}tR and y 0 y 0 y >= 0y \geq 0y0.
a) Find the geodesic equations and the Christoffel symbols.
b) Compute the Riemann curvature tensor and the Ricci scalar.
Problem 2.46 Calculate the Christoffel symbols, the Riemann curvature tensor, the Ricci tensor, and the Ricci scalar for the metric
(2.43) d s 2 = d ρ 2 + ( a 2 + ρ 2 ) d ϕ 2 , (2.43) d s 2 = d ρ 2 + a 2 + ρ 2 d ϕ 2 , {:(2.43)ds^(2)=drho^(2)+(a^(2)+rho^(2))dphi^(2)",":}\begin{equation*} d s^{2}=d \rho^{2}+\left(a^{2}+\rho^{2}\right) d \phi^{2}, \tag{2.43} \end{equation*}(2.43)ds2=dρ2+(a2+ρ2)dϕ2,
where a > 0 a > 0 a > 0a>0a>0 is a constant and the coordinates ρ ρ rho\rhoρ and ϕ ϕ phi\phiϕ vary in the intervals < ρ < < ρ < -oo < rho < oo-\infty<\rho<\infty<ρ< and 0 ϕ < 2 π 0 ϕ < 2 π 0 <= phi < 2pi0 \leq \phi<2 \pi0ϕ<2π, respectively.
Problem 2.47 Show by direct computation of the Riemann curvature tensor that the curvature of the Rindler space with coordinates λ λ lambda\lambdaλ and a a aaa and line element
(2.44) d s 2 = a 2 d λ 2 d a 2 (2.44) d s 2 = a 2 d λ 2 d a 2 {:(2.44)ds^(2)=a^(2)dlambda^(2)-da^(2):}\begin{equation*} d s^{2}=a^{2} d \lambda^{2}-d a^{2} \tag{2.44} \end{equation*}(2.44)ds2=a2dλ2da2
is zero.
Problem 2.48 Consider the curved two-dimensional spacetime t 2 x 2 y 2 = 1 t 2 x 2 y 2 = 1 t^(2)-x^(2)-y^(2)=-1t^{2}-x^{2}-y^{2}=-1t2x2y2=1 embedded in three-dimensional Minkowski spacetime with coordinates ( x μ ) = x μ = (x^(mu))=\left(x^{\mu}\right)=(xμ)= ( t , x , y ) ( t , x , y ) (t,x,y)(t, x, y)(t,x,y) and the metric d s 2 = d t 2 d x 2 d y 2 d s 2 = d t 2 d x 2 d y 2 ds^(2)=dt^(2)-dx^(2)-dy^(2)d s^{2}=d t^{2}-d x^{2}-d y^{2}ds2=dt2dx2dy2. Compute the metric tensor g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν and the Ricci tensor R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν for this two-dimensional spacetime and thus prove that
(2.45) R μ ν = Λ g μ ν (2.45) R μ ν = Λ g μ ν {:(2.45)R_(mu nu)=-Lambdag_(mu nu):}\begin{equation*} R_{\mu \nu}=-\Lambda g_{\mu \nu} \tag{2.45} \end{equation*}(2.45)Rμν=Λgμν
for some constant Λ Λ Lambda\LambdaΛ to be determined. Perform this computation in the coordinate system ( x μ ) = ( λ , φ ) x μ = ( λ , φ ) (x^(mu))=(lambda,varphi)\left(x^{\mu}\right)=(\lambda, \varphi)(xμ)=(λ,φ), where t = sinh ( λ ) , x = cosh ( λ ) cos ( φ ) t = sinh ( λ ) , x = cosh ( λ ) cos ( φ ) t=sinh(lambda),x=cosh(lambda)cos(varphi)t=\sinh (\lambda), x=\cosh (\lambda) \cos (\varphi)t=sinh(λ),x=cosh(λ)cos(φ), and y = y = y=y=y= cosh ( λ ) sin ( φ ) cosh ( λ ) sin ( φ ) cosh(lambda)sin(varphi)\cosh (\lambda) \sin (\varphi)cosh(λ)sin(φ).
Problem 2.49 Compute the Ricci tensor for the two-dimensional spacetime AdS 2 AdS 2 AdS_(2)\mathrm{AdS}_{2}AdS2 and in the coordinates x μ x μ x^(mu)x^{\mu}xμ as defined in Problem 2.27.
Problem 2.50 Consider the 2-dimensional manifold M M MMM defined by being the surface t 2 + u 2 x 2 = α 2 t 2 + u 2 x 2 = α 2 t^(2)+u^(2)-x^(2)=alpha^(2)t^{2}+u^{2}-x^{2}=\alpha^{2}t2+u2x2=α2 (with α > 0 α > 0 alpha > 0\alpha>0α>0 being a constant) embedded in a 3 -dimensional flat manifold with coordinates t , u t , u t,ut, ut,u, and x x xxx, and line element d s 2 = d t 2 + d u 2 d x 2 d s 2 = d t 2 + d u 2 d x 2 ds^(2)=dt^(2)+du^(2)-dx^(2)d s^{2}=d t^{2}+d u^{2}-d x^{2}ds2=dt2+du2dx2.
a) Introduce suitable coordinates on M M MMM and compute the line element for M M MMM in terms of those coordinates.
b) Compute the Christoffel symbols in M M MMM in the coordinates introduced in a).
c) Compute the Ricci scalar in M M MMM.
Problem 2.51 a) Derive the geodesic equations and determine the metric tensor, the Christoffel symbols, the Riemann curvature tensor, the Ricci tensor, and the Ricci scalar for the spherically symmetric metric
2.3 Maxwell's Equations and Energy-Momentum Tensor (2.46) d s 2 = g t t ( t , r ) c 2 d t 2 + g r r ( t , r ) d r 2 r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ,  2.3 Maxwell's Equations and Energy-Momentum Tensor  (2.46) d s 2 = g t t ( t , r ) c 2 d t 2 + g r r ( t , r ) d r 2 r 2 d θ 2 + sin 2 θ d ϕ 2 , {:[" 2.3 Maxwell's Equations and Energy-Momentum Tensor "],[(2.46)ds^(2)=g_(tt)(t","r)c^(2)dt^(2)+g_(rr)(t","r)dr^(2)-r^(2)(dtheta^(2)+sin^(2)theta dphi^(2))","]:}\begin{align*} & \text { 2.3 Maxwell's Equations and Energy-Momentum Tensor } \\ & d s^{2}=g_{t t}(t, r) c^{2} d t^{2}+g_{r r}(t, r) d r^{2}-r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right), \tag{2.46} \end{align*} 2.3 Maxwell's Equations and Energy-Momentum Tensor (2.46)ds2=gtt(t,r)c2dt2+grr(t,r)dr2r2(dθ2+sin2θdϕ2),
using the following parametrization
(2.47) g t t = e v , g r r = e ρ , (2.47) g t t = e v , g r r = e ρ , {:(2.47)g_(tt)=e^(v)","quadg_(rr)=-e^(rho)",":}\begin{equation*} g_{t t}=e^{v}, \quad g_{r r}=-e^{\rho}, \tag{2.47} \end{equation*}(2.47)gtt=ev,grr=eρ,
with the arbitrary functions v = ν ( t , r ) v = ν ( t , r ) v=nu(t,r)v=\nu(t, r)v=ν(t,r) and ρ = ρ ( t , r ) ρ = ρ ( t , r ) rho=rho(t,r)\rho=\rho(t, r)ρ=ρ(t,r).
b) Derive the so-called Schwarzschild solution to Einstein's equations in empty space, i.e., solve G α β = 0 G α β = 0 G_(alpha beta)=0G_{\alpha \beta}=0Gαβ=0 with the spherically symmetric metric given in a).
Hint: Assume that Birkhoff's theorem holds, which states that any spherically symmetric solution to G α β = 0 G α β = 0 G_(alpha beta)=0G_{\alpha \beta}=0Gαβ=0 must be static (and asymptotically flat), i.e., v = 0 v = 0 v^(@)=0\stackrel{\circ}{v}=0v=0 and ρ = 0 ρ = 0 rho^(@)=0\stackrel{\circ}{\rho}=0ρ=0, where a circle denotes partial differentiation with respect to time t t ttt.
Problem 2.52 Prove Birkhoff's theorem, i.e., prove that any spherically symmetric solution to Einstein's equations in empty space must be static.

2.3 Maxwell's Equations and Energy-Momentum Tensor

Problem 2.53 The energy-momentum tensor associated with the electromagnetic field strength tensor F μ v F μ v F^(mu v)F^{\mu v}Fμv is
(2.48) T μ ν = ϵ 0 F λ μ F λ ν + ϵ 0 4 g μ ν F λ ω F λ ω , (2.48) T μ ν = ϵ 0 F λ μ F λ ν + ϵ 0 4 g μ ν F λ ω F λ ω , {:(2.48)T^(mu nu)=epsilon_(0)F_(lambda)^(mu)F^(lambda nu)+(epsilon_(0))/(4)g^(mu nu)F_(lambda omega)F^(lambda omega)",":}\begin{equation*} T^{\mu \nu}=\epsilon_{0} F_{\lambda}^{\mu} F^{\lambda \nu}+\frac{\epsilon_{0}}{4} g^{\mu \nu} F_{\lambda \omega} F^{\lambda \omega}, \tag{2.48} \end{equation*}(2.48)Tμν=ϵ0FλμFλν+ϵ04gμνFλωFλω,
where g μ ν g μ ν g^(mu nu)g^{\mu \nu}gμν is the inverse of the metric tensor g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν. Maxwell's equations in general relativity are written as in Minkowski space, except that partial derivatives are replaced by covariant derivatives, i.e., μ F μ ν = J ν μ F μ ν = J ν grad_(mu)F^(mu nu)=J^(nu)\nabla_{\mu} F^{\mu \nu}=J^{\nu}μFμν=Jν. Show that
(2.49) μ T μ v = ϵ 0 J μ F μ v (2.49) μ T μ v = ϵ 0 J μ F μ v {:(2.49)grad_(mu)T^(mu v)=epsilon_(0)J_(mu)F^(mu v):}\begin{equation*} \nabla_{\mu} T^{\mu v}=\epsilon_{0} J_{\mu} F^{\mu v} \tag{2.49} \end{equation*}(2.49)μTμv=ϵ0JμFμv
Note that this does not violate the relation μ T tot μ ν = 0 μ T tot  μ ν = 0 grad_(mu)T_("tot ")^(mu nu)=0\nabla_{\mu} T_{\text {tot }}^{\mu \nu}=0μTtot μν=0, since the T μ v T μ v T^(mu v)T^{\mu v}Tμv considered in this problem is just the electromagnetic part of the total energy-momentum tensor.
Problem 2.54 Show that half of Maxwell's equations, i.e.,
(2.50) α F β γ + β F γ α + γ F α β = 0 , (2.50) α F β γ + β F γ α + γ F α β = 0 , {:(2.50)del_(alpha)F_(beta gamma)+del_(beta)F_(gamma alpha)+del_(gamma)F_(alpha beta)=0",":}\begin{equation*} \partial_{\alpha} F_{\beta \gamma}+\partial_{\beta} F_{\gamma \alpha}+\partial_{\gamma} F_{\alpha \beta}=0, \tag{2.50} \end{equation*}(2.50)αFβγ+βFγα+γFαβ=0,
can be written precisely in the same form in general relativity; the equations transform covariantly in general coordinate transformations. Why is it unnecessary to write
(2.51) α F β γ + β F γ α + γ F α β = 0 ? (2.51) α F β γ + β F γ α + γ F α β = 0 ? {:(2.51)grad_(alpha)F_(beta gamma)+grad_(beta)F_(gamma alpha)+grad_(gamma)F_(alpha beta)=0?:}\begin{equation*} \nabla_{\alpha} F_{\beta \gamma}+\nabla_{\beta} F_{\gamma \alpha}+\nabla_{\gamma} F_{\alpha \beta}=0 ? \tag{2.51} \end{equation*}(2.51)αFβγ+βFγα+γFαβ=0?
Problem 2.55 Show that the covariant form μ j μ = 0 μ j μ = 0 grad_(mu)j^(mu)=0\nabla_{\mu} j^{\mu}=0μjμ=0 of the current conservation law can be written as g ¯ 1 2 μ ( g ¯ 1 2 j μ ) = 0 g ¯ 1 2 μ g ¯ 1 2 j μ = 0 bar(g)^(-(1)/(2))del_(mu)( bar(g)^((1)/(2))j^(mu))=0\bar{g}^{-\frac{1}{2}} \partial_{\mu}\left(\bar{g}^{\frac{1}{2}} j^{\mu}\right)=0g¯12μ(g¯12jμ)=0, where g ¯ = det ( g μ ν ) ; g μ ν g ¯ = det g μ ν ; g μ ν bar(g)=-det(g_(mu nu));g_(mu nu)\bar{g}=-\operatorname{det}\left(g_{\mu \nu}\right) ; g_{\mu \nu}g¯=det(gμν);gμν is a Lorentzian metric. Show that this is compatible with the generally covariant form μ F μ ν = j ν μ F μ ν = j ν grad_(mu)F^(mu nu)=j^(nu)\nabla_{\mu} F^{\mu \nu}=j^{\nu}μFμν=jν of Maxwell's equations.
Problem 2.56 Assume that in a three-dimensional spacetime, there is a basis of vector fields { X 0 , X 1 , X 2 } X 0 , X 1 , X 2 {X_(0),X_(1),X_(2)}\left\{X_{0}, X_{1}, X_{2}\right\}{X0,X1,X2} with orthogonality relations g ( X μ , X ν ) = 0 g X μ , X ν = 0 g(X_(mu),X_(nu))=0g\left(X_{\mu}, X_{\nu}\right)=0g(Xμ,Xν)=0 for μ v μ v mu!=v\mu \neq vμv and g ( X 0 , X 0 ) = g ( X 1 , X 1 ) = g ( X 2 , X 2 ) = 1 g X 0 , X 0 = g X 1 , X 1 = g X 2 , X 2 = 1 g(X_(0),X_(0))=-g(X_(1),X_(1))=-g(X_(2),X_(2))=1g\left(X_{0}, X_{0}\right)=-g\left(X_{1}, X_{1}\right)=-g\left(X_{2}, X_{2}\right)=1g(X0,X0)=g(X1,X1)=g(X2,X2)=1. In this basis (which is not a coordinate basis!), we define an affine connection by
(2.52) X 0 X 1 = X 1 X 0 = 1 2 X 2 , (2.53) X 1 X 2 = X 2 X 1 = 1 2 X 0 , (2.54) X 2 X 0 = X 0 X 2 = 1 2 X 1 (2.52) X 0 X 1 = X 1 X 0 = 1 2 X 2 , (2.53) X 1 X 2 = X 2 X 1 = 1 2 X 0 , (2.54) X 2 X 0 = X 0 X 2 = 1 2 X 1 {:[(2.52)grad_(X_(0))X_(1)=-grad_(X_(1))X_(0)=(1)/(2)X_(2)","],[(2.53)grad_(X_(1))X_(2)=-grad_(X_(2))X_(1)=-(1)/(2)X_(0)","],[(2.54)grad_(X_(2))X_(0)=-grad_(X_(0))X_(2)=(1)/(2)X_(1)]:}\begin{align*} & \nabla_{X_{0}} X_{1}=-\nabla_{X_{1}} X_{0}=\frac{1}{2} X_{2}, \tag{2.52}\\ & \nabla_{X_{1}} X_{2}=-\nabla_{X_{2}} X_{1}=-\frac{1}{2} X_{0}, \tag{2.53}\\ & \nabla_{X_{2}} X_{0}=-\nabla_{X_{0}} X_{2}=\frac{1}{2} X_{1} \tag{2.54} \end{align*}(2.52)X0X1=X1X0=12X2,(2.53)X1X2=X2X1=12X0,(2.54)X2X0=X0X2=12X1
We also assume that [ X 0 , X 1 ] = X 2 , [ X 1 , X 2 ] = X 0 X 0 , X 1 = X 2 , X 1 , X 2 = X 0 [X_(0),X_(1)]=X_(2),[X_(1),X_(2)]=-X_(0)\left[X_{0}, X_{1}\right]=X_{2},\left[X_{1}, X_{2}\right]=-X_{0}[X0,X1]=X2,[X1,X2]=X0, and [ X 2 , X 0 ] = X 1 X 2 , X 0 = X 1 [X_(2),X_(0)]=X_(1)\left[X_{2}, X_{0}\right]=X_{1}[X2,X0]=X1.
a) Show that grad\nabla is the Levi-Civita connection associated to the metric g g ggg.
b) Compute the energy-momentum tensor T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν corresponding to the metric g g ggg from Einstein's equations in the above basis.
Problem 2.57 The action for a point particle of mass M M MMM in a curved spacetime is given by
(2.55) S M = M g ( γ ˙ , γ ˙ ) d τ (2.55) S M = M g ( γ ˙ , γ ˙ ) d τ {:(2.55)S_(M)=M intsqrt(g(gamma^(˙),gamma^(˙)))d tau:}\begin{equation*} \mathscr{S}_{M}=M \int \sqrt{g(\dot{\gamma}, \dot{\gamma})} d \tau \tag{2.55} \end{equation*}(2.55)SM=Mg(γ˙,γ˙)dτ
where τ τ tau\tauτ is the proper time and γ ˙ γ ˙ gamma^(˙)\dot{\gamma}γ˙ the 4 -velocity of the particle. What is the energymomentum tensor corresponding to such a point particle? Check that your expression takes the expected form in the case of standard coordinates in Minkowski space.
Problem 2.58 The Lagrangian density for an electromagnetic field A μ A μ A_(mu)A_{\mu}Aμ is given by
(2.56) L = 1 4 F μ ν F μ ν + J μ A μ (2.56) L = 1 4 F μ ν F μ ν + J μ A μ {:(2.56)L=-(1)/(4)F_(mu nu)F^(mu nu)+J^(mu)A_(mu):}\begin{equation*} \mathscr{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}+J^{\mu} A_{\mu} \tag{2.56} \end{equation*}(2.56)L=14FμνFμν+JμAμ
where F μ ν = μ A ν ν A μ F μ ν = μ A ν ν A μ F_(mu nu)=del_(mu)A_(nu)-del_(nu)A_(mu)F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}Fμν=μAννAμ and J μ J μ J^(mu)J^{\mu}Jμ is an external 4-current density that does not depend on A μ A μ A_(mu)A_{\mu}Aμ. Use this to derive the equations of motion for an electromagnetic field in a general spacetime.
Problem 2.59 The Lagrangian for a free massive scalar field ϕ ϕ phi\phiϕ in a general spacetime is given by
(2.57) L ϕ = 1 2 [ g μ v ( μ ϕ ) ( ν ϕ ) m 2 ϕ 2 ] . (2.57) L ϕ = 1 2 g μ v μ ϕ ν ϕ m 2 ϕ 2 . {:(2.57)L_(phi)=(1)/(2)[g^(mu v)(del_(mu)phi)(del_(nu)phi)-m^(2)phi^(2)].:}\begin{equation*} \mathcal{L}_{\phi}=\frac{1}{2}\left[g^{\mu v}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \phi\right)-m^{2} \phi^{2}\right] . \tag{2.57} \end{equation*}(2.57)Lϕ=12[gμv(μϕ)(νϕ)m2ϕ2].
Find the equation of motion for ϕ ϕ phi\phiϕ in the curved spacetime based on the principle of stationary action.
Problem 2.60 A massless scalar field ϕ ϕ phi\phiϕ can be described by the Lagrangian density
(2.58) L = 1 2 g μ ν ( μ ϕ ) ( ν ϕ ) V ( ϕ ) , (2.58) L = 1 2 g μ ν μ ϕ ν ϕ V ( ϕ ) , {:(2.58)L=(1)/(2)g^(mu nu)(del_(mu)phi)(del_(nu)phi)-V(phi)",":}\begin{equation*} \mathcal{L}=\frac{1}{2} g^{\mu \nu}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \phi\right)-V(\phi), \tag{2.58} \end{equation*}(2.58)L=12gμν(μϕ)(νϕ)V(ϕ),
where V ( ϕ ) V ( ϕ ) V(phi)V(\phi)V(ϕ) is the potential density, which is a function of ϕ ϕ phi\phiϕ only (i.e., it does not depend on the metric). Compute the components of the stress-energy tensor T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν
and then simplify your expression in the case where g μ ν ( μ ϕ ) ( ν ϕ ) g μ ν μ ϕ ν ϕ g^(mu nu)(del_(mu)phi)(del_(nu)phi)g^{\mu \nu}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \phi\right)gμν(μϕ)(νϕ) is negligible compared to V ( ϕ ) V ( ϕ ) V(phi)V(\phi)V(ϕ).
Problem 2.61 Consider the Robertson-Walker spacetime with metric
(2.59) d s 2 = d t 2 cosh 2 ( H t ) d x 2 (2.59) d s 2 = d t 2 cosh 2 ( H t ) d x 2 {:(2.59)ds^(2)=dt^(2)-cosh^(2)(Ht)dx^(2):}\begin{equation*} d s^{2}=d t^{2}-\cosh ^{2}(H t) d x^{2} \tag{2.59} \end{equation*}(2.59)ds2=dt2cosh2(Ht)dx2
where d x 2 d x 2 dx^(2)d x^{2}dx2 is the standard Euclidean line element in three dimensions. Show that this spacetime is not a vacuum solution to Einstein's field equations.
Problem 2.62 The Lagrangian density of a free electromagnetic field A μ A μ A_(mu)A_{\mu}Aμ is given by
(2.60) L = 1 4 F μ ν F μ ν . (2.60) L = 1 4 F μ ν F μ ν . {:(2.60)L=-(1)/(4)F_(mu nu)F^(mu nu).:}\begin{equation*} \mathscr{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} . \tag{2.60} \end{equation*}(2.60)L=14FμνFμν.
Starting from this Lagrangian density, determine the stress-energy tensor related to the electromagnetic field.

2.4 Killing Vector Fields

Problem 2.63 A Killing vector field X X XXX by definition satisfies the differential equation
(2.61) i X j + j X i = 0 . (2.61) i X j + j X i = 0 . {:(2.61)grad_(i)X_(j)+grad_(j)X_(i)=0.:}\begin{equation*} \nabla_{i} X_{j}+\nabla_{j} X_{i}=0 . \tag{2.61} \end{equation*}(2.61)iXj+jXi=0.
For a geodesic x ( s ) x ( s ) x(s)x(s)x(s), show that there is a conserved quantity
(2.62) W ( s ) = X μ ( x ( s ) ) d x μ d s , (2.62) W ( s ) = X μ ( x ( s ) ) d x μ d s , {:(2.62)W(s)=X_(mu)(x(s))(dx^(mu))/(ds)",":}\begin{equation*} W(s)=X_{\mu}(x(s)) \frac{d x^{\mu}}{d s}, \tag{2.62} \end{equation*}(2.62)W(s)=Xμ(x(s))dxμds,
i.e., the derivative of W W WWW with respect to the curve parameter s s sss vanishes.
Hint: Use the symmetry of the Christoffel symbols from the Levi-Civita connection.
Problem 2.64 The symmetries of a spacetime metric are associated to so-called Killing vector fields. A vector field X is a Killing vector field if L X g = 0 L X g = 0 L_(X)g=0\mathcal{L}_{X} g=0LXg=0; this means that
(2.63) X λ λ g μ ν = g λ ν μ X λ g μ λ ν X λ , (2.63) X λ λ g μ ν = g λ ν μ X λ g μ λ ν X λ , {:(2.63)X^(lambda)del_(lambda)g_(mu nu)=-g_(lambda nu)del_(mu)X^(lambda)-g_(mu lambda)del_(nu)X^(lambda)",":}\begin{equation*} X^{\lambda} \partial_{\lambda} g_{\mu \nu}=-g_{\lambda \nu} \partial_{\mu} X^{\lambda}-g_{\mu \lambda} \partial_{\nu} X^{\lambda}, \tag{2.63} \end{equation*}(2.63)Xλλgμν=gλνμXλgμλνXλ,
for all indices μ , ν μ , ν mu,nu\mu, \nuμ,ν.
a) Show that this condition can be written without reference to any specific choice of local coordinates as
(2.64) X g ( A , B ) = g ( [ X , A ] , B ) + g ( A , [ X , B ] ) , (2.64) X g ( A , B ) = g ( [ X , A ] , B ) + g ( A , [ X , B ] ) , {:(2.64)X*g(A","B)=g([X","A]","B)+g(A","[X","B])",":}\begin{equation*} X \cdot g(A, B)=g([X, A], B)+g(A,[X, B]), \tag{2.64} \end{equation*}(2.64)Xg(A,B)=g([X,A],B)+g(A,[X,B]),
for all vector fields A , B A , B A,BA, BA,B.
b) Show that the coordinate vector field X = λ X = λ X=del_(lambda)X=\partial_{\lambda}X=λ is a Killing vector field if and only if
(2.65) λ g μ ν = 0 , μ , ν (2.65) λ g μ ν = 0 , μ , ν {:(2.65)del_(lambda)g_(mu nu)=0","quad AA mu","nu:}\begin{equation*} \partial_{\lambda} g_{\mu \nu}=0, \quad \forall \mu, \nu \tag{2.65} \end{equation*}(2.65)λgμν=0,μ,ν
c) Show that the vector fields X μ X μ X_(mu)X_{\mu}Xμ in Problem 2.56 are all Killing vector fields.
Problem 2.65 Find the flows of the following vector fields and determine if they are Killing vector fields or not.
a) The field K = y x x y K = y x x y K=ydel_(x)-xdel_(y)K=y \partial_{x}-x \partial_{y}K=yxxy in the Euclidean plane with Cartesian coordinates x x xxx and y y yyy.
b) The field K = x t t x K = x t t x K=xdel_(t)-tdel_(x)K=x \partial_{t}-t \partial_{x}K=xttx in two-dimensional Minkowski space with standard coordinates t t ttt and x x xxx.
Problem 2.66 Consider the paraboloid z = α ( x 2 + y 2 ) z = α x 2 + y 2 z=alpha(x^(2)+y^(2))z=\alpha\left(x^{2}+y^{2}\right)z=α(x2+y2) as a submanifold embedded in Euclidean three-dimensional space ( R 3 ) R 3 (R^(3))\left(\mathbb{R}^{3}\right)(R3).
a) Using coordinates r r rrr and φ φ varphi\varphiφ such that x = r cos ( φ ) x = r cos ( φ ) x=r cos(varphi)x=r \cos (\varphi)x=rcos(φ) and y = r sin ( φ ) y = r sin ( φ ) y=r sin(varphi)y=r \sin (\varphi)y=rsin(φ), compute the induced metric tensor on the paraboloid.
b) Verify that the vector field K = φ K = φ K=del_(varphi)K=\partial_{\varphi}K=φ is a Killing vector field and find an expression for the corresponding conserved quantity for a geodesic in terms of the coordinates and their derivatives along the geodesic.
Problem 2.67 A torus can be parametrized using two angles θ θ theta\thetaθ and φ φ varphi\varphiφ. The metric induced by a typical embedding in R 3 R 3 R^(3)\mathbb{R}^{3}R3 corresponds to the line element
(2.66) d s 2 = [ R + ρ sin ( φ ) ] 2 d θ 2 + ρ 2 d φ 2 . (2.66) d s 2 = [ R + ρ sin ( φ ) ] 2 d θ 2 + ρ 2 d φ 2 . {:(2.66)ds^(2)=[R+rho sin(varphi)]^(2)dtheta^(2)+rho^(2)dvarphi^(2).:}\begin{equation*} d s^{2}=[R+\rho \sin (\varphi)]^{2} d \theta^{2}+\rho^{2} d \varphi^{2} . \tag{2.66} \end{equation*}(2.66)ds2=[R+ρsin(φ)]2dθ2+ρ2dφ2.
a) Find the Christoffel symbols corresponding to the Levi-Civita connection of this metric.
b) Find a Killing vector field for the torus and the corresponding conserved quantity along geodesics of the Levi-Civita connection.
c) It is possible to introduce a flat connection ~ ~ tilde(grad)\tilde{\nabla}~ with all connection coefficients Γ ~ a b c = 0 Γ ~ a b c = 0 tilde(Gamma)_(ab)^(c)=0\tilde{\Gamma}_{a b}^{c}=0Γ~abc=0 in the θ φ θ φ theta-varphi\theta-\varphiθφ coordinate system. This connection is not metric compatible. Compute the components of the derivative ~ a g ~ a g tilde(grad)_(a)g\tilde{\nabla}_{a} g~ag for this connection.
Problem 2.68 A wavy two-dimensional surface locally described by the coordinates ρ ρ rho\rhoρ and φ φ varphi\varphiφ is embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3 according to
(2.67) x = ρ cos ( φ ) , y = ρ sin ( φ ) , z = R 0 cos ( ρ / R 0 ) , (2.67) x = ρ cos ( φ ) , y = ρ sin ( φ ) , z = R 0 cos ρ / R 0 , {:(2.67)x=rho cos(varphi)","quad y=rho sin(varphi)","quad z=R_(0)cos(rho//R_(0))",":}\begin{equation*} x=\rho \cos (\varphi), \quad y=\rho \sin (\varphi), \quad z=R_{0} \cos \left(\rho / R_{0}\right), \tag{2.67} \end{equation*}(2.67)x=ρcos(φ),y=ρsin(φ),z=R0cos(ρ/R0),
where R 0 > 0 R 0 > 0 R_(0) > 0R_{0}>0R0>0 is a constant.
a) Compute the induced metric tensor and the Christoffel symbols.
b) Find the flow for each of the following vector fields and determine whether or not they are Killing vector fields:
(2.68) K = ρ , Q = φ . (2.68) K = ρ , Q = φ . {:(2.68)K=del_(rho)","quad Q=del_(varphi).:}\begin{equation*} K=\partial_{\rho}, \quad Q=\partial_{\varphi} . \tag{2.68} \end{equation*}(2.68)K=ρ,Q=φ.
Problem 2.69 The 2-dimensional de Sitter space dS 2 dS 2 dS_(2)\mathrm{dS}_{2}dS2 may be defined as the surface t 2 x 2 y 2 = r 0 2 t 2 x 2 y 2 = r 0 2 t^(2)-x^(2)-y^(2)=-r_(0)^(2)t^{2}-x^{2}-y^{2}=-r_{0}^{2}t2x2y2=r02 in 1+2-dimensional Minkowski space, where r 0 > 0 r 0 > 0 r_(0) > 0r_{0}>0r0>0 is a constant.
a) Introduce suitable coordinates on dS 2 dS 2 dS_(2)\mathrm{dS}_{2}dS2.
b) Compute the components of the metric tensor induced by the embedding in Minkowski space in your selected coordinates.
c) Find (at least) two Killing vector fields on dS 2 dS 2 dS_(2)\mathrm{dS}_{2}dS2.

2.5 Schwarzschild Metric

Problem 2.70 The Schwarzschild metric, when restricted to the plane θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2, is given by
(2.69) d s 2 = ( 1 α r ) ( d x 0 ) 2 ( 1 α r ) 1 d r 2 r 2 d ϕ 2 (2.69) d s 2 = 1 α r d x 0 2 1 α r 1 d r 2 r 2 d ϕ 2 {:(2.69)ds^(2)=(1-(alpha )/(r))(dx^(0))^(2)-(1-(alpha )/(r))^(-1)dr^(2)-r^(2)dphi^(2):}\begin{equation*} d s^{2}=\left(1-\frac{\alpha}{r}\right)\left(d x^{0}\right)^{2}-\left(1-\frac{\alpha}{r}\right)^{-1} d r^{2}-r^{2} d \phi^{2} \tag{2.69} \end{equation*}(2.69)ds2=(1αr)(dx0)2(1αr)1dr2r2dϕ2
Derive the geodesic equations of motion for a test particle in this metric.
Problem 2.71 The Schwarzschild metric is normally written in terms of time and spherical coordinates. Transform this metric to coordinates ( x 1 , x 2 , x 3 ) = x 1 , x 2 , x 3 = (x^(1),x^(2),x^(3))=\left(x^{1}, x^{2}, x^{3}\right)=(x1,x2,x3)= r ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) r ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) r(sin theta cos phi,sin theta sin phi,cos theta)r(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)r(sinθcosϕ,sinθsinϕ,cosθ).
Problem 2.72 Consider the Schwarzschild metric
d s 2 = c 2 d τ 2 = ( 1 2 G M c 2 r ) ( d x 0 ) 2 ( 1 2 G M c 2 r ) 1 d r 2 r 2 d Ω 2 d s 2 = c 2 d τ 2 = 1 2 G M c 2 r d x 0 2 1 2 G M c 2 r 1 d r 2 r 2 d Ω 2 ds^(2)=c^(2)dtau^(2)=(1-(2GM)/(c^(2)r))(dx^(0))^(2)-(1-(2GM)/(c^(2)r))^(-1)dr^(2)-r^(2)dOmega^(2)d s^{2}=c^{2} d \tau^{2}=\left(1-\frac{2 G M}{c^{2} r}\right)\left(d x^{0}\right)^{2}-\left(1-\frac{2 G M}{c^{2} r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2}ds2=c2dτ2=(12GMc2r)(dx0)2(12GMc2r)1dr2r2dΩ2,
where τ τ tau\tauτ is the proper time, x 0 = c t x 0 = c t x^(0)=ctx^{0}=c tx0=ct, and d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}dΩ2=dθ2+sin2θdϕ2.
a) Assuming circular motion in the equatorial plane, i.e., r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0, where r 0 r 0 r_(0)r_{0}r0 is a constant, derive Kepler's third law
(2.71) Δ t = 2 π r 0 3 G M (2.71) Δ t = 2 π r 0 3 G M {:(2.71)Delta t=2pisqrt((r_(0)^(3))/(GM)):}\begin{equation*} \Delta t=2 \pi \sqrt{\frac{r_{0}^{3}}{G M}} \tag{2.71} \end{equation*}(2.71)Δt=2πr03GM
where Δ t Δ t Delta t\Delta tΔt is the period. Compare with the classical result.
b) Compute the proper time Δ τ Δ τ Delta tau\Delta \tauΔτ for one period of circular motion.
Problem 2.73 The Schwarzschild metric is given by
(2.72) d s 2 = ( 1 r r ) ( d x 0 ) 2 ( 1 r r ) 1 d r 2 r 2 d Ω 2 (2.72) d s 2 = 1 r r d x 0 2 1 r r 1 d r 2 r 2 d Ω 2 {:(2.72)ds^(2)=(1-(r_(**))/(r))(dx^(0))^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right)\left(d x^{0}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.72} \end{equation*}(2.72)ds2=(1rr)(dx0)2(1rr)1dr2r2dΩ2
where d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}dΩ2=dθ2+sin2θdϕ2 and r 2 G M / c 2 r 2 G M / c 2 r_(**)-=2GM//c^(2)r_{*} \equiv 2 G M / c^{2}r2GM/c2 is the Schwarzschild radius. Find the worldlines for bodies, outside of the Schwarzschild horizon, radially freely falling toward the black hole.
Problem 2.74 Show that there are no circular free fall orbits inside of the radius r = 3 r / 2 r = 3 r / 2 r=3r_(**)//2r=3 r_{*} / 2r=3r/2 in the Scwharzschild spacetime.
Problem 2.75 For the Schwarzschild solution in the limit r r r r r≫r_(**)r \gg r_{*}rr and approximately circular orbits such that r = r 0 + ρ r = r 0 + ρ r=r_(0)+rhor=r_{0}+\rhor=r0+ρ, where ρ r 0 ρ r 0 rho≪r_(0)\rho \ll r_{0}ρr0, determine the ratio between the period of oscillations in ρ ρ rho\rhoρ to the orbital period.
Problem 2.76 The optical size of a black hole is given by 4 π b 2 4 π b 2 4pib^(2)4 \pi b^{2}4πb2, where b b bbb is the minimal impact parameter such that the past-null geodesics originate at r r r rarr oor \rightarrow \inftyr.
Find the optical size of the Schwarzschild black hole for which the line element is given by
(2.73) d s 2 = ( 1 R r ) d t 2 ( 1 R r ) 1 d r 2 r 2 d Ω 2 (2.73) d s 2 = 1 R r d t 2 1 R r 1 d r 2 r 2 d Ω 2 {:(2.73)ds^(2)=(1-(R)/(r))dt^(2)-(1-(R)/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{R}{r}\right) d t^{2}-\left(1-\frac{R}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.73} \end{equation*}(2.73)ds2=(1Rr)dt2(1Rr)1dr2r2dΩ2
Hint: For null geodesics in the Schwarzschild spacetime, the angular momentum is equal to the impact parameter (i.e., b = L b = L b=Lb=Lb=L ) for r ˙ = 1 r ˙ = 1 r^(˙)=1\dot{r}=1r˙=1 at r r r rarr oor \rightarrow \inftyr.
Problem 2.77 Show that the Schwarzschild solution of Einstein's equation
(2.74) d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 [ d θ 2 + sin ( θ ) 2 d φ 2 ] , (2.74) d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d θ 2 + sin ( θ ) 2 d φ 2 , {:(2.74)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)[dtheta^(2)+sin(theta)^(2)dvarphi^(2)]",":}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2}\left[d \theta^{2}+\sin (\theta)^{2} d \varphi^{2}\right], \tag{2.74} \end{equation*}(2.74)ds2=(1rr)dt2(1rr)1dr2r2[dθ2+sin(θ)2dφ2],
where r = 2 G M r = 2 G M r_(**)=2GMr_{*}=2 G Mr=2GM and c = 1 c = 1 c=1c=1c=1, can be embedded in 5+1-dimensional Minkowski spacetime with coordinates ( Z 1 , Z 2 , , Z 6 ) Z 1 , Z 2 , , Z 6 (Z_(1),Z_(2),dots,Z_(6))\left(Z_{1}, Z_{2}, \ldots, Z_{6}\right)(Z1,Z2,,Z6) and metric
(2.75) d s 2 = d Z 1 2 d Z 2 2 d Z 3 2 d Z 4 2 d Z 5 2 d Z 6 2 (2.75) d s 2 = d Z 1 2 d Z 2 2 d Z 3 2 d Z 4 2 d Z 5 2 d Z 6 2 {:(2.75)ds^(2)=dZ_(1)^(2)-dZ_(2)^(2)-dZ_(3)^(2)-dZ_(4)^(2)-dZ_(5)^(2)-dZ_(6)^(2):}\begin{equation*} d s^{2}=d Z_{1}^{2}-d Z_{2}^{2}-d Z_{3}^{2}-d Z_{4}^{2}-d Z_{5}^{2}-d Z_{6}^{2} \tag{2.75} \end{equation*}(2.75)ds2=dZ12dZ22dZ32dZ42dZ52dZ62
Hint: Make the ansatz Z 1 = 2 r sinh ( t / ( 2 r ) ) f ( r ) , Z 2 = 2 r cosh ( t / ( 2 r ) ) f ( r ) Z 1 = 2 r sinh t / 2 r f ( r ) , Z 2 = 2 r cosh t / 2 r f ( r ) Z_(1)=2r_(**)sinh(t//(2r_(**)))f(r),Z_(2)=2r_(**)cosh(t//(2r_(**)))f(r)Z_{1}=2 r_{*} \sinh \left(t /\left(2 r_{*}\right)\right) f(r), Z_{2}=2 r_{*} \cosh \left(t /\left(2 r_{*}\right)\right) f(r)Z1=2rsinh(t/(2r))f(r),Z2=2rcosh(t/(2r))f(r), Z 3 = g ( r ) , Z 4 = r sin ( θ ) cos ( φ ) , Z 5 = r sin ( θ ) sin ( φ ) Z 3 = g ( r ) , Z 4 = r sin ( θ ) cos ( φ ) , Z 5 = r sin ( θ ) sin ( φ ) Z_(3)=g(r),Z_(4)=r sin(theta)cos(varphi),Z_(5)=r sin(theta)sin(varphi)Z_{3}=g(r), Z_{4}=r \sin (\theta) \cos (\varphi), Z_{5}=r \sin (\theta) \sin (\varphi)Z3=g(r),Z4=rsin(θ)cos(φ),Z5=rsin(θ)sin(φ), and Z 6 = r cos ( θ ) Z 6 = r cos ( θ ) Z_(6)=r cos(theta)Z_{6}=r \cos (\theta)Z6=rcos(θ) and determine the functions f ( r ) f ( r ) f(r)f(r)f(r) and g ( r ) g ( r ) g(r)g(r)g(r) [see C. Fronsdal, Phys. Rev. 116, 778 (1959)]. The answer may contain an integral.

2.6 Metrics, Geodesic Equations, and Proper Quantities

Problem 2.78 The restriction of the Minkowski metric ( η μ ν ) η μ ν (eta_(mu nu))\left(\eta_{\mu \nu}\right)(ημν) to the threedimensional hyperboloid M 3 M 3 M_(3)M_{3}M3, i.e.,
(2.76) ( x 0 ) 2 ( x 1 ) 2 ( x 2 ) 2 ( x 3 ) 2 = a 2 (2.76) x 0 2 x 1 2 x 2 2 x 3 2 = a 2 {:(2.76)(x^(0))^(2)-(x^(1))^(2)-(x^(2))^(2)-(x^(3))^(2)=-a^(2):}\begin{equation*} \left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}=-a^{2} \tag{2.76} \end{equation*}(2.76)(x0)2(x1)2(x2)2(x3)2=a2
defines a curved metric on M 3 M 3 M_(3)M_{3}M3. Determine the lightlike geodesics with constant spherical angle ϕ ϕ phi\phiϕ (where x 1 = r sin θ cos ϕ , x 2 = r sin θ sin ϕ x 1 = r sin θ cos ϕ , x 2 = r sin θ sin ϕ x^(1)=r sin theta cos phi,x^(2)=r sin theta sin phix^{1}=r \sin \theta \cos \phi, x^{2}=r \sin \theta \sin \phix1=rsinθcosϕ,x2=rsinθsinϕ, and x 3 = r cos θ x 3 = r cos θ x^(3)=r cos thetax^{3}=r \cos \thetax3=rcosθ as usual).
Problem 2.79 The Minkowski metric d s 2 = ( d x 0 ) 2 ( d x 1 ) 2 ( d x 2 ) 2 d s 2 = d x 0 2 d x 1 2 d x 2 2 ds^(2)=(dx^(0))^(2)-(dx^(1))^(2)-(dx^(2))^(2)d s^{2}=\left(d x^{0}\right)^{2}-\left(d x^{1}\right)^{2}-\left(d x^{2}\right)^{2}ds2=(dx0)2(dx1)2(dx2)2 in R 3 R 3 R^(3)\mathbb{R}^{3}R3 induces a nonflat Lorentzian metric on the surface S = { ( x 0 , x 1 , x 2 ) : ( x 0 ) 2 ( x 1 ) 2 S = x 0 , x 1 , x 2 : x 0 2 x 1 2 S={(x^(0),x^(1),x^(2)):(x^(0))^(2)-(x^(1))^(2)-:}S=\left\{\left(x^{0}, x^{1}, x^{2}\right):\left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\right.S={(x0,x1,x2):(x0)2(x1)2 ( x 2 ) 2 = 1 } x 2 2 = 1 {:(x^(2))^(2)=-1}\left.\left(x^{2}\right)^{2}=-1\right\}(x2)2=1}. Let ϕ ϕ phi\phiϕ be the polar angle in the ( x 1 , x 2 ) x 1 , x 2 (x^(1),x^(2))\left(x^{1}, x^{2}\right)(x1,x2)-plane. Compute the global time difference Δ x 0 Δ x 0 Deltax^(0)\Delta x^{0}Δx0 needed for a light signal to travel from a point ϕ 0 = 0 ϕ 0 = 0 phi_(0)=0\phi_{0}=0ϕ0=0 to a point ϕ = π / 2 ϕ = π / 2 phi=pi//2\phi=\pi / 2ϕ=π/2 on S S SSS.
Problem 2.80 Let ( x 0 ( s ) , r ( s ) , θ ( s ) , ϕ ( s ) ) x 0 ( s ) , r ( s ) , θ ( s ) , ϕ ( s ) (x^(0)(s),r(s),theta(s),phi(s))\left(x^{0}(s), r(s), \theta(s), \phi(s)\right)(x0(s),r(s),θ(s),ϕ(s)) be a lightlike geodesic for the Schwarzschild metric, expressed in the spherical coordinates ( r , θ , ϕ ) ( r , θ , ϕ ) (r,theta,phi)(r, \theta, \phi)(r,θ,ϕ). Derive a differential equation for r ( s ) r ( s ) r(s)r(s)r(s) in the form
(2.77) d r d s = f ( r ) (2.77) d r d s = f ( r ) {:(2.77)(dr)/(ds)=f(r):}\begin{equation*} \frac{d r}{d s}=f(r) \tag{2.77} \end{equation*}(2.77)drds=f(r)
when restricted to the plane θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2.
Hint: The following nonzero Christoffel symbols for the Schwarzschild metric when θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2 might be useful: Γ 0 r 0 = Γ r r r = 1 2 α d α d r , Γ 00 r = α 2 d α d r , Γ θ θ r = Γ ϕ ϕ r = r α Γ 0 r 0 = Γ r r r = 1 2 α d α d r , Γ 00 r = α 2 d α d r , Γ θ θ r = Γ ϕ ϕ r = r α Gamma_(0r)^(0)=-Gamma_(rr)^(r)=(1)/(2alpha)(d alpha)/(dr),Gamma_(00)^(r)=(alpha)/(2)(d alpha)/(dr),Gamma_(theta theta)^(r)=Gamma_(phi phi)^(r)=-r alpha\Gamma_{0 r}^{0}=-\Gamma_{r r}^{r}=\frac{1}{2 \alpha} \frac{d \alpha}{d r}, \Gamma_{00}^{r}=\frac{\alpha}{2} \frac{d \alpha}{d r}, \Gamma_{\theta \theta}^{r}=\Gamma_{\phi \phi}^{r}=-r \alphaΓ0r0=Γrrr=12αdαdr,Γ00r=α2dαdr,Γθθr=Γϕϕr=rα, and Γ r θ θ = Γ r ϕ ϕ = 1 r Γ r θ θ = Γ r ϕ ϕ = 1 r Gamma_(r theta)^(theta)=Gamma_(r phi)^(phi)=(1)/(r)\Gamma_{r \theta}^{\theta}=\Gamma_{r \phi}^{\phi}=\frac{1}{r}Γrθθ=Γrϕϕ=1r, where α = α ( r ) = 1 2 G M c 2 r α = α ( r ) = 1 2 G M c 2 r alpha=alpha(r)=1-(2GM)/(c^(2)r)\alpha=\alpha(r)=1-\frac{2 G M}{c^{2} r}α=α(r)=12GMc2r.
Problem 2.81 Consider the metric d s 2 = c 2 d t 2 S ( t ) 2 ( d x 2 + d y 2 + d z 2 ) d s 2 = c 2 d t 2 S ( t ) 2 d x 2 + d y 2 + d z 2 ds^(2)=c^(2)dt^(2)-S(t)^(2)(dx^(2)+dy^(2)+dz^(2))d s^{2}=c^{2} d t^{2}-S(t)^{2}\left(d x^{2}+d y^{2}+d z^{2}\right)ds2=c2dt2S(t)2(dx2+dy2+dz2), where S ( t ) S ( t ) S(t)S(t)S(t) is an increasing function of time t t ttt with S ( 0 ) = 0 S ( 0 ) = 0 S(0)=0S(0)=0S(0)=0. Find the geodesic equations of motion. In particular, construct explicitly the lightlike geodesic when S ( t ) = t / t 0 S ( t ) = t / t 0 S(t)=t//t_(0)S(t)=t / t_{0}S(t)=t/t0 for some constant t 0 > 0 t 0 > 0 t_(0) > 0t_{0}>0t0>0. What are the points ( c t , x , y , z ) R 4 ( c t , x , y , z ) R 4 (ct,x,y,z)inR^(4)(c t, x, y, z) \in \mathbb{R}^{4}(ct,x,y,z)R4 for a fixed t > t 0 t > t 0 t > t_(0)t>t_{0}t>t0, which are causally related to the event p = ( c t 0 , c t 0 , 0 , 0 ) p = c t 0 , c t 0 , 0 , 0 p=(ct_(0),ct_(0),0,0)p=\left(c t_{0}, c t_{0}, 0,0\right)p=(ct0,ct0,0,0), i.e., the points which are connected to p p ppp by a future-directed timelike (or lightlike) curve?
Problem 2.82 The line element on the unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2 is given by
(2.78) d s 2 = d θ 2 + sin 2 ( θ ) d φ 2 , (2.78) d s 2 = d θ 2 + sin 2 ( θ ) d φ 2 , {:(2.78)ds^(2)=dtheta^(2)+sin^(2)(theta)dvarphi^(2)",":}\begin{equation*} d s^{2}=d \theta^{2}+\sin ^{2}(\theta) d \varphi^{2}, \tag{2.78} \end{equation*}(2.78)ds2=dθ2+sin2(θ)dφ2,
and the nonzero Christoffel symbols are
(2.79) Γ φ φ θ = sin ( θ ) cos ( θ ) , Γ θ φ φ = Γ φ θ φ = cot ( θ ) . (2.79) Γ φ φ θ = sin ( θ ) cos ( θ ) , Γ θ φ φ = Γ φ θ φ = cot ( θ ) . {:(2.79)Gamma_(varphi varphi)^(theta)=-sin(theta)cos(theta)","quadGamma_(theta varphi)^(varphi)=Gamma_(varphi theta)^(varphi)=cot(theta).:}\begin{equation*} \Gamma_{\varphi \varphi}^{\theta}=-\sin (\theta) \cos (\theta), \quad \Gamma_{\theta \varphi}^{\varphi}=\Gamma_{\varphi \theta}^{\varphi}=\cot (\theta) . \tag{2.79} \end{equation*}(2.79)Γφφθ=sin(θ)cos(θ),Γθφφ=Γφθφ=cot(θ).
Consider two geodesics separated by a small distance δ δ delta\deltaδ and both orthogonal to the equator θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2, compute the rate of acceleration of the geodesics toward each other through the geodesic deviation equation
(2.80) A a = R b c d a χ ˙ b χ ˙ c X d , (2.80) A a = R b c d a χ ˙ b χ ˙ c X d , {:(2.80)A^(a)=R_(bcd)^(a)chi^(˙)^(b)chi^(˙)^(c)X^(d)",":}\begin{equation*} A^{a}=R_{b c d}^{a} \dot{\chi}^{b} \dot{\chi}^{c} X^{d}, \tag{2.80} \end{equation*}(2.80)Aa=Rbcdaχ˙bχ˙cXd,
where X d X d X^(d)X^{d}Xd is the infinitesimal separation of the geodesics at the equator and χ a χ a chi^(a)\chi^{a}χa are the coordinates.
Problem 2.83 The metric for the de Sitter universe can be expressed in the form
(2.81) d s 2 = d t 2 e 2 t / R ( d x 2 + d y 2 + d z 2 ) , (2.81) d s 2 = d t 2 e 2 t / R d x 2 + d y 2 + d z 2 , {:(2.81)ds^(2)=dt^(2)-e^(2t//R)(dx^(2)+dy^(2)+dz^(2))",":}\begin{equation*} d s^{2}=d t^{2}-e^{2 t / R}\left(d x^{2}+d y^{2}+d z^{2}\right), \tag{2.81} \end{equation*}(2.81)ds2=dt2e2t/R(dx2+dy2+dz2),
where R > 0 R > 0 R > 0R>0R>0 is a constant and x , y x , y x,yx, yx,y, and z z zzz can be treated as rectangular coordinates and t t ttt as time.
a) Show that the trajectories of freely falling particles and photons are straight lines.
b) A body at a point x = X > 0 x = X > 0 x=X > 0x=X>0x=X>0 on the x x xxx-axis emits a photon toward the origin at time t = 0 t = 0 t=0t=0t=0. Show that, if X < R X < R X < RX<RX<R, the photon arrives at the origin at t = R log ( 1 X / R ) t = R log ( 1 X / R ) t=-R log(1-X//R)t=-R \log (1-X / R)t=Rlog(1X/R).
Problem 2.84 The de Sitter universe dS 4 dS 4 dS_(4)\mathrm{dS}_{4}dS4 is defined as the hyperboloid
(2.82) t 2 ( x 1 ) 2 ( x 2 ) 2 ( x 3 ) 2 ( x 4 ) 2 = T 0 2 , (2.82) t 2 x 1 2 x 2 2 x 3 2 x 4 2 = T 0 2 , {:(2.82)t^(2)-(x^(1))^(2)-(x^(2))^(2)-(x^(3))^(2)-(x^(4))^(2)=-T_(0)^(2)",":}\begin{equation*} t^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}-\left(x^{4}\right)^{2}=-T_{0}^{2}, \tag{2.82} \end{equation*}(2.82)t2(x1)2(x2)2(x3)2(x4)2=T02,
in units with c = 1 c = 1 c=1c=1c=1 and where T 0 > 0 T 0 > 0 T_(0) > 0T_{0}>0T0>0 is a constant. The metric on dS 4 dS 4 dS_(4)\mathrm{dS}_{4}dS4 is defined by restricting the five-dimensional Minkowski metric to the hyperboloid.
a) Find an explicit expression for the four-dimensional metric in a suitable coordinate system on dS 4 dS 4 dS_(4)\mathrm{dS}_{4}dS4.
b) Derive the geodesic equations in dS 4 dS 4 dS_(4)\mathrm{dS}_{4}dS4.
c) Compare the metric tensor with the Robertson-Walker metric
(2.83) d s 2 = d t 2 S ( t ) 2 d Ω 2 , (2.83) d s 2 = d t 2 S ( t ) 2 d Ω 2 , {:(2.83)ds^(2)=dt^(2)-S(t)^(2)dOmega^(2)",":}\begin{equation*} d s^{2}=d t^{2}-S(t)^{2} d \Omega^{2}, \tag{2.83} \end{equation*}(2.83)ds2=dt2S(t)2dΩ2,
by writing down your metric in a coordinate system where the coefficient in front of the timelike coordinate is identically equal to one.
Problem 2.85 Consider AdS 2 AdS 2 AdS_(2)\mathrm{AdS}_{2}AdS2 and the coordinates x μ x μ x^(mu)x^{\mu}xμ defined in Problem 2.27.
a) Find the trajectory of a light ray in this spacetime.
b) A particle at rest in r = r 0 > 2 r = r 0 > 2 r=r_(0) > 2r=r_{0}>2r=r0>2 starts to fall freely at t = 0 t = 0 t=0t=0t=0. What is the proper time it takes for the particle to freely fall and reach r = r 1 r = r 1 r=r_(1)r=r_{1}r=r1, where 1 < r 1 < r 0 1 < r 1 < r 0 1 < r_(1) < r_(0)1<r_{1}<r_{0}1<r1<r0 ? Also compute the coordinate time for this fall, i.e., the time t t ttt at which the particle reaches r 1 r 1 r_(1)r_{1}r1. Discuss your result in the limit r 1 1 r 1 1 r_(1)rarr1r_{1} \rightarrow 1r11, and in particular, a possible physical interpretation of what you find. (Your answers may contain integrals and functions defined by implicit equations.)
Problem 2.86 Four-dimensional anti-de Sitter space, which is also called AdS 4 AdS 4 AdS_(4)\mathrm{AdS}_{4}AdS4, is a four-dimensional curved space that can be defined as follows: Consider the fivedimensional space with coordinates ( X a ) = ( U , V , X , Y , Z ) X a = ( U , V , X , Y , Z ) (X^(a))=(U,V,X,Y,Z)\left(X^{a}\right)=(U, V, X, Y, Z)(Xa)=(U,V,X,Y,Z), where a = 1 , 2 , 3 , 4 , 5 a = 1 , 2 , 3 , 4 , 5 a=1,2,3,4,5a=1,2,3,4,5a=1,2,3,4,5, and metric
(2.84) d s 2 = d U 2 + d V 2 d X 2 d Y 2 d Z 2 d X a d X a (2.84) d s 2 = d U 2 + d V 2 d X 2 d Y 2 d Z 2 d X a d X a {:(2.84)ds^(2)=dU^(2)+dV^(2)-dX^(2)-dY^(2)-dZ^(2)-=dX^(a)dX_(a):}\begin{equation*} d s^{2}=d U^{2}+d V^{2}-d X^{2}-d Y^{2}-d Z^{2} \equiv d X^{a} d X_{a} \tag{2.84} \end{equation*}(2.84)ds2=dU2+dV2dX2dY2dZ2dXadXa
Then, AdS 4 AdS 4 AdS_(4)\mathrm{AdS}_{4}AdS4 is the subspace of this space defined by the relation
(2.85) X a X a = U 2 + V 2 X 2 Y 2 Z 2 = 1 (2.85) X a X a = U 2 + V 2 X 2 Y 2 Z 2 = 1 {:(2.85)X^(a)X_(a)=U^(2)+V^(2)-X^(2)-Y^(2)-Z^(2)=1:}\begin{equation*} X^{a} X_{a}=U^{2}+V^{2}-X^{2}-Y^{2}-Z^{2}=1 \tag{2.85} \end{equation*}(2.85)XaXa=U2+V2X2Y2Z2=1
Hint: Note that ( X a ) = ( U , V , X , Y , Z ) X a = ( U , V , X , Y , Z ) (X_(a))=(U,V,-X,-Y,-Z)\left(X_{a}\right)=(U, V,-X,-Y,-Z)(Xa)=(U,V,X,Y,Z).
a) Compute the metric of AdS 4 AdS 4 AdS_(4)\operatorname{AdS}_{4}AdS4 in the coordinate system ( x μ ) = ( α , λ , θ , φ ) x μ = ( α , λ , θ , φ ) (x^(mu))=(alpha,lambda,theta,varphi)\left(x^{\mu}\right)=(\alpha, \lambda, \theta, \varphi)(xμ)=(α,λ,θ,φ), where μ = 0 , 1 , 2 , 3 μ = 0 , 1 , 2 , 3 mu=0,1,2,3\mu=0,1,2,3μ=0,1,2,3, defined as follows: Let ( r , θ , φ ) ( r , θ , φ ) (r,theta,varphi)(r, \theta, \varphi)(r,θ,φ) be spherical coordinates for ( X , Y , Z ) ( X , Y , Z ) (X,Y,Z)(X, Y, Z)(X,Y,Z), i.e., X = r sin ( θ ) cos ( φ ) , Y = r sin ( θ ) sin ( φ ) X = r sin ( θ ) cos ( φ ) , Y = r sin ( θ ) sin ( φ ) X=r sin(theta)cos(varphi),Y=r sin(theta)sin(varphi)X=r \sin (\theta) \cos (\varphi), Y=r \sin (\theta) \sin (\varphi)X=rsin(θ)cos(φ),Y=rsin(θ)sin(φ), and Z = r cos ( θ ) Z = r cos ( θ ) Z=r cos(theta)Z=r \cos (\theta)Z=rcos(θ), and ( t , α ) ( t , α ) (t,alpha)(t, \alpha)(t,α) polar coordinates for ( U , V ) ( U , V ) (U,V)(U, V)(U,V), i.e., U = t cos ( α ) U = t cos ( α ) U=t cos(alpha)U=t \cos (\alpha)U=tcos(α) and V = t sin ( α ) V = t sin ( α ) V=t sin(alpha)V=t \sin (\alpha)V=tsin(α). Then, t = cosh ( λ ) t = cosh ( λ ) t=cosh(lambda)t=\cosh (\lambda)t=cosh(λ) and r = sinh ( λ ) r = sinh ( λ ) r=sinh(lambda)r=\sinh (\lambda)r=sinh(λ).
b) Compute all possible trajectories λ ( α ) λ ( α ) lambda(alpha)\lambda(\alpha)λ(α) of light rays on AdS 4 AdS 4 AdS_(4)\mathrm{AdS}_{4}AdS4 moving along the subspace Y = Z = 0 Y = Z = 0 Y=Z=0Y=Z=0Y=Z=0.
c) Prove that all lightlike geodesics on AdS 4 AdS 4 AdS_(4)\mathrm{AdS}_{4}AdS4 are straight lines in the embedding space, i.e., they obey the equations
(2.86) X ¨ a = 0 , (2.86) X ¨ a = 0 , {:(2.86)X^(¨)^(a)=0",":}\begin{equation*} \ddot{X}^{a}=0, \tag{2.86} \end{equation*}(2.86)X¨a=0,
with the dot indicating differentiation with respect to proper time τ τ tau\tauτ.
Hint: You can find these trajectories by extremizing the functional L d τ L d τ intLd tau\int \mathcal{L} d \tauLdτ with
(2.87) L = 1 2 X ˙ a X ˙ a + λ ( X a X a 1 ) , (2.87) L = 1 2 X ˙ a X ˙ a + λ X a X a 1 , {:(2.87)L=(1)/(2)X^(˙)^(a)X^(˙)_(a)+lambda(X^(a)X_(a)-1)",":}\begin{equation*} \mathcal{L}=\frac{1}{2} \dot{X}^{a} \dot{X}_{a}+\lambda\left(X^{a} X_{a}-1\right), \tag{2.87} \end{equation*}(2.87)L=12X˙aX˙a+λ(XaXa1),
and λ λ lambda\lambdaλ a Lagrange multiplier (explain why this is so!). One key step in the proof is to show that X ˙ a X ˙ a X ˙ a X ˙ a X^(˙)^(a)X^(˙)_(a)\dot{X}^{a} \dot{X}_{a}X˙aX˙a is conserved and you can assume that it is identically equal to zero for a lightlike trajectory.
Problem 2.87 Consider the 1+1-dimensional Robertson-Walker spacetime described by the metric
(2.88) d s 2 = d t 2 a ( t ) 2 d x 2 (2.88) d s 2 = d t 2 a ( t ) 2 d x 2 {:(2.88)ds^(2)=dt^(2)-a(t)^(2)dx^(2):}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2} d x^{2} \tag{2.88} \end{equation*}(2.88)ds2=dt2a(t)2dx2
for some function a ( t ) a ( t ) a(t)a(t)a(t).
a) Compute the Ricci tensor R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν for this spacetime.
b) Derive the trajectory x ( t ) x ( t ) x(t)x(t)x(t) of a light ray on this spacetime for a ( t ) = 1 / ( A 2 + a ( t ) = 1 / A 2 + a(t)=1//(A^(2)+:}a(t)=1 /\left(A^{2}+\right.a(t)=1/(A2+ B 2 t 2 B 2 t 2 B^(2)t^(2)B^{2} t^{2}B2t2 ) and some constants A A AAA and B B BBB. Assume that the light ray is emitted at t = 0 t = 0 t=0t=0t=0 from x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0 with d x / d t > 0 d x / d t > 0 dx//dt > 0d x / d t>0dx/dt>0.
Problem 2.88 Consider the 1+2-dimensional Robertson-Walker spacetime described by the metric
(2.89) d s 2 = d t 2 a ( t ) 2 ( d r 2 1 k r 2 + r 2 d ϕ 2 ) (2.89) d s 2 = d t 2 a ( t ) 2 d r 2 1 k r 2 + r 2 d ϕ 2 {:(2.89)ds^(2)=dt^(2)-a(t)^(2)((dr^(2))/(1-kr^(2))+r^(2)dphi^(2)):}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2}\left(\frac{d r^{2}}{1-k r^{2}}+r^{2} d \phi^{2}\right) \tag{2.89} \end{equation*}(2.89)ds2=dt2a(t)2(dr21kr2+r2dϕ2)
where a ( t ) a ( t ) a(t)a(t)a(t) is some function and k k kkk is a constant. Derive the geodesic equations and determine the Christoffel symbols.
Problem 2.89 The Robertson-Walker metric describing a particular closed universe is given by
(2.90) d s 2 = d t 2 e 2 t / a [ 1 1 + ( r / a ) 2 d r 2 + r 2 ( sin 2 θ d φ 2 + d θ 2 ) ] , (2.90) d s 2 = d t 2 e 2 t / a 1 1 + ( r / a ) 2 d r 2 + r 2 sin 2 θ d φ 2 + d θ 2 , {:(2.90)ds^(2)=dt^(2)-e^(-2t//a)[(1)/(1+(r//a)^(2))dr^(2)+r^(2)(sin^(2)theta dvarphi^(2)+dtheta^(2))]",":}\begin{equation*} d s^{2}=d t^{2}-e^{-2 t / a}\left[\frac{1}{1+(r / a)^{2}} d r^{2}+r^{2}\left(\sin ^{2} \theta d \varphi^{2}+d \theta^{2}\right)\right], \tag{2.90} \end{equation*}(2.90)ds2=dt2e2t/a[11+(r/a)2dr2+r2(sin2θdφ2+dθ2)],
for some parameter a > 0 a > 0 a > 0a>0a>0.
a) Determine all nonzero Christoffel symbols Γ μ ν r , μ , ν = t , r , θ , φ Γ μ ν r , μ , ν = t , r , θ , φ Gamma_(mu nu)^(r),mu,nu=t,r,theta,varphi\Gamma_{\mu \nu}^{r}, \mu, \nu=t, r, \theta, \varphiΓμνr,μ,ν=t,r,θ,φ for this metric. Moreover, for the vector field with the components A t = t / a , A r = r / a , A θ = A t = t / a , A r = r / a , A θ = A^(t)=t//a,A^(r)=r//a,A^(theta)=A^{t}=t / a, A^{r}=r / a, A^{\theta}=At=t/a,Ar=r/a,Aθ= A φ = 0 A φ = 0 A^(varphi)=0A^{\varphi}=0Aφ=0 in the coordinate vector basis, compute the components μ A r , μ = t , r μ A r , μ = t , r grad_(mu)A^(r),mu=t,r\nabla_{\mu} A^{r}, \mu=t, rμAr,μ=t,r, of the covariant derivative.
b) Find the trajectory r ( t ) r ( t ) r(t)r(t)r(t) of a light pulse emitted at time t = 0 t = 0 t=0t=0t=0 at r = a r = a r=ar=ar=a and moving radially outward such that θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and φ = 0 φ = 0 varphi=0\varphi=0φ=0 at all times.
Problem 2.90 The Robertson-Walker metric is defined by
(2.91) d s 2 = c 2 d t 2 S ( t ) 2 ( d r 2 1 k r 2 + r 2 d Ω 2 ) (2.91) d s 2 = c 2 d t 2 S ( t ) 2 d r 2 1 k r 2 + r 2 d Ω 2 {:(2.91)ds^(2)=c^(2)dt^(2)-S(t)^(2)((dr^(2))/(1-kr^(2))+r^(2)dOmega^(2)):}\begin{equation*} d s^{2}=c^{2} d t^{2}-S(t)^{2}\left(\frac{d r^{2}}{1-k r^{2}}+r^{2} d \Omega^{2}\right) \tag{2.91} \end{equation*}(2.91)ds2=c2dt2S(t)2(dr21kr2+r2dΩ2)
for some smooth function S ( t ) S ( t ) S(t)S(t)S(t) and d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}dΩ2=dθ2+sin2θdϕ2. We consider the case k = 1 k = 1 k=1k=1k=1. After a coordinate transformation χ = arcsin r χ = arcsin r chi=arcsin r\chi=\arcsin rχ=arcsinr (with 0 χ π / 2 0 χ π / 2 0 <= chi <= pi//20 \leq \chi \leq \pi / 20χπ/2 ), this can be written as
(2.92) d s 2 = c 2 d t 2 S ( t ) 2 ( d χ 2 + sin 2 χ d Ω 2 ) (2.92) d s 2 = c 2 d t 2 S ( t ) 2 d χ 2 + sin 2 χ d Ω 2 {:(2.92)ds^(2)=c^(2)dt^(2)-S(t)^(2)(dchi^(2)+sin^(2)chi dOmega^(2)):}\begin{equation*} d s^{2}=c^{2} d t^{2}-S(t)^{2}\left(d \chi^{2}+\sin ^{2} \chi d \Omega^{2}\right) \tag{2.92} \end{equation*}(2.92)ds2=c2dt2S(t)2(dχ2+sin2χdΩ2)
a) Derive first integrals for the geodesic equations when d Ω = 0 d Ω = 0 d Omega=0d \Omega=0dΩ=0.
b) Derive a formula expressing the distance a light ray emitted at r = 0 r = 0 r=0r=0r=0 at universal time t 0 t 0 t_(0)t_{0}t0 travels (in the r r rrr coordinate) in the time interval [ t 0 , t 0 + T ] t 0 , t 0 + T [t_(0),t_(0)+T]\left[t_{0}, t_{0}+T\right][t0,t0+T].
Problem 2.91 A spaceship is freely falling (along a geodesic) toward the true singularity at r = 0 r = 0 r=0r=0r=0 in a Schwarzschild black hole. The initial velocity is θ ˙ = ϕ ˙ = 0 θ ˙ = ϕ ˙ = 0 theta^(˙)=phi^(˙)=0\dot{\theta}=\dot{\phi}=0θ˙=ϕ˙=0, r ˙ = α r ˙ = α r^(˙)=alpha\dot{r}=\alphar˙=α, and t ˙ = β t ˙ = β t^(˙)=beta\dot{t}=\betat˙=β, where the dot means differentiation with respect to the path
parameter (which can be taken to be the proper time) and the standard Schwarzschild metric is used, i.e.,
(2.93) d s 2 = ( 1 2 G M c 2 r ) c 2 d t 2 ( 1 2 G M c 2 r ) 1 d r 2 r 2 d Ω 2 (2.93) d s 2 = 1 2 G M c 2 r c 2 d t 2 1 2 G M c 2 r 1 d r 2 r 2 d Ω 2 {:(2.93)ds^(2)=(1-(2GM)/(c^(2)r))c^(2)dt^(2)-(1-(2GM)/(c^(2)r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{2 G M}{c^{2} r}\right) c^{2} d t^{2}-\left(1-\frac{2 G M}{c^{2} r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.93} \end{equation*}(2.93)ds2=(12GMc2r)c2dt2(12GMc2r)1dr2r2dΩ2
The proper time τ τ tau\tauτ needed to reach the singularity r = 0 r = 0 r=0r=0r=0, when starting from r = r = r=r=r= r 0 < 2 G M / c 2 r 0 < 2 G M / c 2 r_(0) < 2GM//c^(2)r_{0}<2 G M / c^{2}r0<2GM/c2, can be written as
(2.94) τ = 0 r 0 f ( r ) d r (2.94) τ = 0 r 0 f ( r ) d r {:(2.94)tau=int_(0)^(r_(0))f(r)dr:}\begin{equation*} \tau=\int_{0}^{r_{0}} f(r) d r \tag{2.94} \end{equation*}(2.94)τ=0r0f(r)dr
What is the function f ( r ) f ( r ) f(r)f(r)f(r) ?
Problem 2.92 A Schwarzschild black hole has a mass M = 13.5 10 30 kg M = 13.5 10 30 kg M=13.5*10^(30)kgM=13.5 \cdot 10^{30} \mathrm{~kg}M=13.51030 kg (about seven times the solar mass). An observer is freely falling (along a geodesic) radially toward the black hole. The initial radial coordinate is r = r 0 = 10 10 km r = r 0 = 10 10 km r=r_(0)=10^(10)kmr=r_{0}=10^{10} \mathrm{~km}r=r0=1010 km and the initial coordinate velocity is c ( d r / d x 0 ) = v 0 = 10 km / s c d r / d x 0 = v 0 = 10 km / s c(dr//dx^(0))=-v_(0)=-10km//sc\left(d r / d x^{0}\right)=-v_{0}=-10 \mathrm{~km} / \mathrm{s}c(dr/dx0)=v0=10 km/s. Derive the formula for the proper time needed to reach the Schwarzschild horizon and give the order of magnitude of this time. Newton's gravitational constant is G 6.67 10 11 m 3 / ( kg s 2 ) G 6.67 10 11 m 3 / kg s 2 G~~6.67*10^(-11)m^(3)//(kg*s^(2))G \approx 6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right)G6.671011 m3/(kgs2) and the speed of light is c 3 10 8 m / s c 3 10 8 m / s c~~3*10^(8)m//sc \approx 3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}c3108 m/s.
Hint: The following integral can be useful
d x a + b x = 1 a [ x 2 + b x a b 2 a ln ( x + b 2 a + x 2 + b x a ) ] + C d x a + b x = 1 a x 2 + b x a b 2 a ln x + b 2 a + x 2 + b x a + C int(dx)/(sqrt(a+(b)/(x)))=(1)/(sqrta)[sqrt(x^(2)+(bx)/(a))-(b)/(2a)ln(x+(b)/(2a)+sqrt(x^(2)+(bx)/(a)))]+C\int \frac{d x}{\sqrt{a+\frac{b}{x}}}=\frac{1}{\sqrt{a}}\left[\sqrt{x^{2}+\frac{b x}{a}}-\frac{b}{2 a} \ln \left(x+\frac{b}{2 a}+\sqrt{x^{2}+\frac{b x}{a}}\right)\right]+Cdxa+bx=1a[x2+bxab2aln(x+b2a+x2+bxa)]+C,
where a , b a , b a,ba, ba,b, and C C CCC are constants.
Problem 2.93 A particle of mass m > 0 m > 0 m > 0m>0m>0 is freely falling radially toward the horizon of a Schwarzschild black hole of mass M M MMM. Show that p 0 = m c g 00 x ˙ 0 p 0 = m c g 00 x ˙ 0 p_(0)=mcg_(00)x^(˙)^(0)p_{0}=m c g_{00} \dot{x}^{0}p0=mcg00x˙0 is a constant of motion. Find the proper time Δ s Δ s Delta s\Delta sΔs (as a function of p 0 = E / c p 0 = E / c p_(0)=E//cp_{0}=E / cp0=E/c ) needed for the particle to reach r = 2 G M / c 2 r = 2 G M / c 2 r=2GM//c^(2)r=2 G M / c^{2}r=2GM/c2 from r = 3 G M / c 2 r = 3 G M / c 2 r=3GM//c^(2)r=3 G M / c^{2}r=3GM/c2. Show that the result can be written as
(2.96) Δ s = r 3 r / 2 d r ( E m c 2 ) 2 ( 1 r r ) (2.96) Δ s = r 3 r / 2 d r E m c 2 2 1 r r {:(2.96)Delta s=int_(r_(**))^(3r_(**)//2)(dr)/(sqrt(((E)/(mc^(2)))^(2)-(1-(r_(**))/(r)))):}\begin{equation*} \Delta s=\int_{r_{*}}^{3 r_{*} / 2} \frac{d r}{\sqrt{\left(\frac{E}{m c^{2}}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)}} \tag{2.96} \end{equation*}(2.96)Δs=r3r/2dr(Emc2)2(1rr)
where r 2 G M / c 2 r 2 G M / c 2 r_(**)-=2GM//c^(2)r_{*} \equiv 2 G M / c^{2}r2GM/c2.
Problem 2.94 An observer in the Schwarzschild spacetime moves with fixed radial coordinate r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 and fixed angular velocity φ ˙ = ω φ ˙ = ω varphi^(˙)=omega\dot{\varphi}=\omegaφ˙=ω in the plane θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2. Compute the 4 -acceleration A A AAA and the proper acceleration α α alpha\alphaα of the observer as a function of the proper period ω ω omega\omegaω.
Problem 2.95 A particle of mass m > 0 m > 0 m > 0m>0m>0 is freely falling radially toward the event horizon r = 2 G M r = 2 G M r=2GMr=2 G Mr=2GM of a Schwarzschild black hole of mass M M MMM (we set c = 1 c = 1 c=1c=1c=1 ), i.e., θ θ theta\thetaθ and φ φ varphi\varphiφ are constant in the standard coordinates where the metric is
d s 2 = ( 1 2 G M r ) d t 2 ( 1 2 G M r ) 1 d r 2 r 2 [ d θ 2 + sin ( θ ) 2 d φ 2 ] d s 2 = 1 2 G M r d t 2 1 2 G M r 1 d r 2 r 2 d θ 2 + sin ( θ ) 2 d φ 2 ds^(2)=(1-(2GM)/(r))dt^(2)-(1-(2GM)/(r))^(-1)dr^(2)-r^(2)[dtheta^(2)+sin(theta)^(2)dvarphi^(2)]d s^{2}=\left(1-\frac{2 G M}{r}\right) d t^{2}-\left(1-\frac{2 G M}{r}\right)^{-1} d r^{2}-r^{2}\left[d \theta^{2}+\sin (\theta)^{2} d \varphi^{2}\right]ds2=(12GMr)dt2(12GMr)1dr2r2[dθ2+sin(θ)2dφ2].
a) Compute the trajectory of this particle.
b) Find also formulas for the coordinate time Δ t Δ t Delta t\Delta tΔt and proper time Δ τ Δ τ Delta tau\Delta \tauΔτ it takes for the particle to reach the event horizon from some position r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0. Determine for each of these times if it is finite or infinite, and discuss the physical significance of your results.
Integration constants can be fixed so that the results have a simple form, but if so a physical interpretation of what the choice amounts to should be given.
Problem 2.96 Consider an observer in the Schwarzschild spacetime with line element
(2.98) d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 d Ω 2 (2.98) d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d Ω 2 {:(2.98)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.98} \end{equation*}(2.98)ds2=(1rr)dt2(1rr)1dr2r2dΩ2
The observer starts out at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 and initially moves tangentially with a local velocity v 0 v 0 v_(0)v_{0}v0 relative to the stationary frame.
a) Determine the minimal value of v 0 v 0 v_(0)v_{0}v0 such that the observer does not fall into the black hole region of the solution.
b) Assuming that v 0 = 0 v 0 = 0 v_(0)=0v_{0}=0v0=0, compute the proper time it takes for the observer to reach the singularity.
Hint: The local velocity v v vvv of an observer relative to the stationary frame has a γ γ gamma\gammaγ factor of γ = V U γ = V U gamma=V*U\gamma=V \cdot Uγ=VU, where V V VVV is the 4-velocity of the observer, U = α t U = α t U=alphadel_(t)U=\alpha \partial_{t}U=αt, and U 2 = 1 U 2 = 1 U^(2)=1U^{2}=1U2=1.
Problem 2.97 You are sending up a satellite around the Earth. You want it to be directed such that when you turn off its engines, it will follow a geodesic around the Earth with fixed radius. The metric around the Earth is
d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 d θ 2 r 2 sin 2 θ d ϕ 2 , R 0 > r d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d θ 2 r 2 sin 2 θ d ϕ 2 , R 0 > r ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dtheta^(2)-r^(2)sin^(2)theta dphi^(2),quadR_(0) > r_(**)d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \theta^{2}-r^{2} \sin ^{2} \theta d \phi^{2}, \quad R_{0}>r_{*}ds2=(1rr)dt2(1rr)1dr2r2dθ2r2sin2θdϕ2,R0>r,
where R 0 R 0 R_(0)R_{0}R0 is the radius of the Earth. Your initial data when you turn off the engines at τ = 0 τ = 0 tau=0\tau=0τ=0 are the following
(2.100) d r d τ | τ = 0 = 0 , d θ d τ | τ = 0 = 0 , d ϕ d τ | τ = 0 = B , r | τ = 0 = R , θ | τ = 0 = π / 2 , ϕ | τ = 0 = 0 (2.100) d r d τ τ = 0 = 0 , d θ d τ τ = 0 = 0 , d ϕ d τ τ = 0 = B , r τ = 0 = R , θ τ = 0 = π / 2 , ϕ τ = 0 = 0 {:[(2.100)(dr)/(d tau)|_(tau=0)=0"," quad(d theta)/(d tau)|_(tau=0)=0"," quad(d phi)/(d tau)|_(tau=0)=B","],[r|_(tau=0)=R"," quad theta|_(tau=0)=pi//2"," quad phi|_(tau=0)=0]:}\begin{align*} & \left.\frac{d r}{d \tau}\right|_{\tau=0}=0,\left.\quad \frac{d \theta}{d \tau}\right|_{\tau=0}=0,\left.\quad \frac{d \phi}{d \tau}\right|_{\tau=0}=B, \tag{2.100}\\ & \left.r\right|_{\tau=0}=R,\left.\quad \theta\right|_{\tau=0}=\pi / 2,\left.\quad \phi\right|_{\tau=0}=0 \end{align*}(2.100)drdτ|τ=0=0,dθdτ|τ=0=0,dϕdτ|τ=0=B,r|τ=0=R,θ|τ=0=π/2,ϕ|τ=0=0
Is this possible? If so, determine how your initial condition B B BBB, which you have to choose, depends on R R RRR and r r r_(**)r_{*}r.
Problem 2.98 A satellite moves at a constant radial distance from the Earth with a constant orbital coordinate speed v = r d ϕ / d t v = r d ϕ / d t v=rd phi//dtv=r d \phi / d tv=rdϕ/dt. Assume that the metric is the Schwarzschild metric and let the orbit be in the plane with angle θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 such that
(2.101) d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 d ϕ 2 (2.101) d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d ϕ 2 {:(2.101)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dphi^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \phi^{2} \tag{2.101} \end{equation*}(2.101)ds2=(1rr)dt2(1rr)1dr2r2dϕ2
where r 2 G M r 2 G M r_(**)-=2GMr_{*} \equiv 2 G Mr2GM is the Schwarzschild radius.
a) Calculate the proper time τ τ tau\tauτ for the satellite to complete one orbit around the Earth. Express the answer in terms of the coordinate speed v v vvv and the radius r r rrr.
b) Use the result in a) to calculate t / τ t / τ t//taut / \taut/τ and show that if this is series expanded to first order in v v vvv and the gravitational potential, it holds that
(2.102) t τ 1 v 2 2 Φ s (2.102) t τ 1 v 2 2 Φ s {:(2.102)(t)/( tau)-1≃(v^(2))/(2)-Phi_(s):}\begin{equation*} \frac{t}{\tau}-1 \simeq \frac{v^{2}}{2}-\Phi_{s} \tag{2.102} \end{equation*}(2.102)tτ1v22Φs
where Φ s Φ s Phi_(s)\Phi_{s}Φs is the gravitational potential at the satellite.
Problem 2.99 The spacetime outside of the Earth may be approximately described by the Schwarzschild line element
(2.103) d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 d Ω 2 (2.103) d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d Ω 2 {:(2.103)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.103} \end{equation*}(2.103)ds2=(1rr)dt2(1rr)1dr2r2dΩ2
where r r r_(**)r_{*}r is the Schwarzschild radius of the Earth (approximately 9 mm ). A GPS satellite is orbiting the Earth in free fall at a stationary radius r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0. The motion is assumed to occur in the plane θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2.
a) Since r r rrr is constant, the motion will have a 4 -velocity U = α t + β φ U = α t + β φ U=alphadel_(t)+betadel_(varphi)U=\alpha \partial_{t}+\beta \partial_{\varphi}U=αt+βφ. Find the values of the constants α α alpha\alphaα and β β beta\betaβ.
b) Find an expression for the proper time it takes for the satellite to complete a full orbit around the Earth.
c) An observer is stationary at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 (note that this requires proper acceleration of this observer). At what speed will the satellite pass by the observer?
Hint: The relative speed v v vvv between two objects with 4 -velocities U U UUU and V V VVV, respectively, has a γ γ gamma\gammaγ factor of γ = V U γ = V U gamma=V*U\gamma=V \cdot Uγ=VU.
Problem 2.100 Consider two observers in the exterior Schwarzschild spacetime with line element
(2.104) d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 d Ω 2 (2.104) d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d Ω 2 {:(2.104)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.104} \end{equation*}(2.104)ds2=(1rr)dt2(1rr)1dr2r2dΩ2
Both observers can be assumed to move in the plane θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2. The first observer is a stationary observer with fixed spatial coordinates r = r 0 > 3 r r = r 0 > 3 r r=r_(0) > 3r_(**)r=r_{0}>3 r_{*}r=r0>3r and φ = φ 0 φ = φ 0 varphi=varphi_(0)\varphi=\varphi_{0}φ=φ0, whereas the second observer is moving on a circular geodesic with radius r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0.
a) Give a parametrization of the worldline for each observer and use it to find the proper acceleration of the observers.
b) The observers meet and synchronize their clocks when they pass each other. Find the ratio between the times shown by the respective clocks when they pass each other the next time.
c) Find the relative velocity of the observers as they pass each other.
Problem 2.101 You have reached a fast rotating neutron star with your spaceship. You decide that you want to go around the neutron star once. Let your orbit be at constant radial distance and your coordinate speed v = r d ϕ / d t v = r d ϕ / d t v=rd phi//dtv=r d \phi / d tv=rdϕ/dt.
a) Calculate the proper time τ τ tau\tauτ it takes you to go around. Express your answer in terms of the radius R R RRR and the speed v v vvv (set c = 1 c = 1 c=1c=1c=1 ). The metric describing this neutron star is given by the Kerr metric, i.e.,
d s 2 = ( 1 r r ρ 2 ) d t 2 + 2 a r r sin 2 θ ρ 2 d t d ϕ ρ 2 Δ d r 2 ρ 2 d θ 2 (2.105) ( r 2 + a 2 + a 2 r r sin 2 θ ρ 2 ) sin 2 θ d ϕ 2 d s 2 = 1 r r ρ 2 d t 2 + 2 a r r sin 2 θ ρ 2 d t d ϕ ρ 2 Δ d r 2 ρ 2 d θ 2 (2.105) r 2 + a 2 + a 2 r r sin 2 θ ρ 2 sin 2 θ d ϕ 2 {:[ds^(2)=(1-(rr_(**))/(rho^(2)))dt^(2)+(2arr_(**)sin^(2)theta)/(rho^(2))dtd phi-(rho^(2))/(Delta)dr^(2)-rho^(2)dtheta^(2)],[(2.105)-(r^(2)+a^(2)+(a^(2)rr_(**)sin^(2)theta)/(rho^(2)))sin^(2)theta dphi^(2)]:}\begin{align*} d s^{2}= & \left(1-\frac{r r_{*}}{\rho^{2}}\right) d t^{2}+\frac{2 a r r_{*} \sin ^{2} \theta}{\rho^{2}} d t d \phi-\frac{\rho^{2}}{\Delta} d r^{2}-\rho^{2} d \theta^{2} \\ & -\left(r^{2}+a^{2}+\frac{a^{2} r r_{*} \sin ^{2} \theta}{\rho^{2}}\right) \sin ^{2} \theta d \phi^{2} \tag{2.105} \end{align*}ds2=(1rrρ2)dt2+2arrsin2θρ2dtdϕρ2Δdr2ρ2dθ2(2.105)(r2+a2+a2rrsin2θρ2)sin2θdϕ2
where ρ 2 r 2 + a 2 cos 2 θ , Δ r 2 r r + a 2 , a J / M ρ 2 r 2 + a 2 cos 2 θ , Δ r 2 r r + a 2 , a J / M rho^(2)-=r^(2)+a^(2)cos^(2)theta,Delta-=r^(2)-rr_(**)+a^(2),a-=J//M\rho^{2} \equiv r^{2}+a^{2} \cos ^{2} \theta, \Delta \equiv r^{2}-r r_{*}+a^{2}, a \equiv J / Mρ2r2+a2cos2θ,Δr2rr+a2,aJ/M, and M M MMM and J J JJJ are the mass and the angular momentum of the neutron star, respectively. You choose to make the orbit at a fixed angle of θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2.
b) Now, use your result in a) to calculate T / τ T / τ T//tauT / \tauT/τ, where T T TTT is the coordinate time of your orbit, and show if you expand to first approximation in v v vvv and the gravitational potential, you obtain
(2.106) T τ 1 + v 2 2 [ 1 + ( a R ) 2 ] + r 2 R , if r R v 2 (2.106) T τ 1 + v 2 2 1 + a R 2 + r 2 R ,  if  r R v 2 {:(2.106)(T)/( tau)≃1+(v^(2))/(2)[1+((a)/(R))^(2)]+(r_(**))/(2R)","quad" if "(r_(**))/(R)∼v^(2):}\begin{equation*} \frac{T}{\tau} \simeq 1+\frac{v^{2}}{2}\left[1+\left(\frac{a}{R}\right)^{2}\right]+\frac{r_{*}}{2 R}, \quad \text { if } \frac{r_{*}}{R} \sim v^{2} \tag{2.106} \end{equation*}(2.106)Tτ1+v22[1+(aR)2]+r2R, if rRv2
Problem 2.102 At the time of inflation, consider a massive free-falling particle. Assume that the metric of spacetime is
(2.107) d s 2 = d t 2 a ( t ) 2 [ d ρ 2 + ρ 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] , a ( t ) = a 0 e C t , (2.107) d s 2 = d t 2 a ( t ) 2 d ρ 2 + ρ 2 d θ 2 + sin 2 θ d ϕ 2 , a ( t ) = a 0 e C t , {:(2.107)ds^(2)=dt^(2)-a(t)^(2)[drho^(2)+rho^(2)(dtheta^(2)+sin^(2)theta dphi^(2))]","quad a(t)=a_(0)e^(Ct)",":}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2}\left[d \rho^{2}+\rho^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right], \quad a(t)=a_{0} e^{C t}, \tag{2.107} \end{equation*}(2.107)ds2=dt2a(t)2[dρ2+ρ2(dθ2+sin2θdϕ2)],a(t)=a0eCt,
where a 0 a 0 a_(0)a_{0}a0 and C C CCC are constants. The initial values for the free-falling particle are given by
(2.108) d ρ d τ | τ = 0 = 1 4 , d θ d τ | τ = 0 = 1 4 , d ϕ d τ | τ = 0 = 1 4 t | τ = 0 = 0 , ρ | τ = 0 = 1 , θ | τ = 0 = 0 , ϕ | τ = 0 = 0 (2.108) d ρ d τ τ = 0 = 1 4 , d θ d τ τ = 0 = 1 4 , d ϕ d τ τ = 0 = 1 4 t τ = 0 = 0 , ρ τ = 0 = 1 , θ τ = 0 = 0 , ϕ τ = 0 = 0 {:[(2.108)(d rho)/(d tau)|_(tau=0)=(1)/(4)"," quad(d theta)/(d tau)|_(tau=0)=(1)/(4)"," quad(d phi)/(d tau)|_(tau=0)=(1)/(4)],[t|_(tau=0)=0"," quad rho|_(tau=0)=1"," quad theta|_(tau=0)=0"," quad phi|_(tau=0)=0]:}\begin{align*} & \left.\frac{d \rho}{d \tau}\right|_{\tau=0}=\frac{1}{4},\left.\quad \frac{d \theta}{d \tau}\right|_{\tau=0}=\frac{1}{4},\left.\quad \frac{d \phi}{d \tau}\right|_{\tau=0}=\frac{1}{4} \tag{2.108}\\ & \left.t\right|_{\tau=0}=0,\left.\quad \rho\right|_{\tau=0}=1,\left.\quad \theta\right|_{\tau=0}=0,\left.\quad \phi\right|_{\tau=0}=0 \end{align*}(2.108)dρdτ|τ=0=14,dθdτ|τ=0=14,dϕdτ|τ=0=14t|τ=0=0,ρ|τ=0=1,θ|τ=0=0,ϕ|τ=0=0
a) Is the spatial geometry curved or not (at each fixed value of time)? Justify your answer.
b) Calculate the proper time for the free-falling particle between the coordinate times t = 0 t = 0 t=0t=0t=0 and t = t 1 t = t 1 t=t_(1)t=t_{1}t=t1.
Hint: Calculations might become simpler if another coordinate system is used! It is also okay to give the answer expressed as an integral.
Problem 2.103 Assume a three-dimensional version of the Robertson-Walker metric (with k = 0 k = 0 k=0k=0k=0 ):
(2.109) d s 2 = d t 2 a ( t ) 2 ( d r 2 + r 2 d ϕ 2 ) (2.109) d s 2 = d t 2 a ( t ) 2 d r 2 + r 2 d ϕ 2 {:(2.109)ds^(2)=dt^(2)-a(t)^(2)(dr^(2)+r^(2)dphi^(2)):}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2}\left(d r^{2}+r^{2} d \phi^{2}\right) \tag{2.109} \end{equation*}(2.109)ds2=dt2a(t)2(dr2+r2dϕ2)
You are traveling in your spaceship in this universe. You decide to travel in a circle around r = 0 r = 0 r=0r=0r=0 on a fixed radius r = R 0 r = R 0 r=R_(0)r=R_{0}r=R0.
a) Calculate the proper time for the spaceship in this geometry going around one time (let ϕ ϕ phi\phiϕ go from zero to 2 π 2 π 2pi2 \pi2π ) if you have the following constant velocity v = a ( t ) R 0 d ϕ d t v = a ( t ) R 0 d ϕ d t v=a(t)R_(0)(d phi)/(dt)v=a(t) R_{0} \frac{d \phi}{d t}v=a(t)R0dϕdt and a ( t ) = exp ( t ) a ( t ) = exp ( t ) a(t)=exp(t)a(t)=\exp (t)a(t)=exp(t). Let the initial time be t = 0 t = 0 t=0t=0t=0.
b) Is it always possible to get around in a finite time?
Problem 2.104 You are traveling in your spaceship in outer intergalactic space. The metric can be assumed to be the Robertson-Walker metric with zero curvature:
(2.110) d s 2 = d t 2 a ( t ) 2 ( d x 2 + d y 2 + d z 2 ) (2.110) d s 2 = d t 2 a ( t ) 2 d x 2 + d y 2 + d z 2 {:(2.110)ds^(2)=dt^(2)-a(t)^(2)(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2}\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{2.110} \end{equation*}(2.110)ds2=dt2a(t)2(dx2+dy2+dz2)
To save fuel, you do not use the spaceship's engines. You are moving according to the following initial conditions
(2.111) d x d τ | τ = 0 = A , d y d τ | τ = 0 = 0 , d z d τ | τ = 0 = 0 , x | τ = 0 = X 0 , y | τ = 0 = 0 , z | τ = 0 = 0 (2.111) d x d τ τ = 0 = A , d y d τ τ = 0 = 0 , d z d τ τ = 0 = 0 , x τ = 0 = X 0 , y τ = 0 = 0 , z τ = 0 = 0 {:[(2.111)(dx)/(d tau)|_(tau=0)=A"," quad(dy)/(d tau)|_(tau=0)=0"," quad(dz)/(d tau)|_(tau=0)=0","],[x|_(tau=0)=X_(0)"," quad y|_(tau=0)=0"," quad z|_(tau=0)=0]:}\begin{align*} & \left.\frac{d x}{d \tau}\right|_{\tau=0}=A,\left.\quad \frac{d y}{d \tau}\right|_{\tau=0}=0,\left.\quad \frac{d z}{d \tau}\right|_{\tau=0}=0, \tag{2.111}\\ & \left.x\right|_{\tau=0}=X_{0},\left.\quad y\right|_{\tau=0}=0,\left.\quad z\right|_{\tau=0}=0 \end{align*}(2.111)dxdτ|τ=0=A,dydτ|τ=0=0,dzdτ|τ=0=0,x|τ=0=X0,y|τ=0=0,z|τ=0=0
Calculate the proper time it takes you to reach x = X D x = X D x=X_(D)x=X_{D}x=XD. It is sufficient to give your answer in terms of an integral, which depends on the initial value A A AAA and the scale factor a ( t ) a ( t ) a(t)a(t)a(t).
Problem 2.105 An observer ( A ) ( A ) (A)(A)(A) is stationary in Schwarzschild spacetime at a radius r 0 r 0 r_(0)r_{0}r0. A second observer ( B ) ( B ) (B)(B)(B) is initially stationary at r = r 1 r = r 1 r=r_(1)r=r_{1}r=r1, but at some event (which can be assigned t = τ = 0 t = τ = 0 t=tau=0t=\tau=0t=τ=0 ) suffers from a rocket failure and becomes freely falling. On the way into the black hole, B B BBB passes right by A A AAA.
a) What is the relative velocity of A A AAA and B B BBB as they pass by each other?
b) What proper time has passed for B B BBB since the rocket failure when they pass by each other?
Problem 2.106 For the two-dimensional spacetime with coordinates t t ttt and x x xxx and line element d s 2 = x 2 d t 2 d x 2 d s 2 = x 2 d t 2 d x 2 ds^(2)=x^(2)dt^(2)-dx^(2)d s^{2}=x^{2} d t^{2}-d x^{2}ds2=x2dt2dx2 :
a) Compute the proper acceleration of an observer with worldline x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0, where x 0 x 0 x_(0)x_{0}x0 is a constant.
b) Compute the proper time for a free-falling observer starting at x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0 with d x / d t = 0 d x / d t = 0 dx//dt=0d x / d t=0dx/dt=0 to reach the coordinate singularity at x = 0 x = 0 x=0x=0x=0.

2.7 Kruskal-Szekeres Coordinates

Problem 2.107 Show that for r > 2 μ r > 2 μ r > 2mur>2 \mur>2μ the Kruskal-Szekeres metric
(2.112) d s 2 = 16 μ 2 r e ( 2 μ r ) / 2 μ d u d v r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (2.112) d s 2 = 16 μ 2 r e ( 2 μ r ) / 2 μ d u d v r 2 d θ 2 + sin 2 θ d ϕ 2 {:(2.112)ds^(2)=(16mu^(2))/(r)e^((2mu-r)//2mu)dudv-r^(2)(dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*} d s^{2}=\frac{16 \mu^{2}}{r} e^{(2 \mu-r) / 2 \mu} d u d v-r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{2.112} \end{equation*}(2.112)ds2=16μ2re(2μr)/2μdudvr2(dθ2+sin2θdϕ2)
is equivalent to the standard Schwarzschild metric through the relations
(2.113) u v = ( 2 μ r ) e ( r 2 μ ) / 2 μ , t = x 0 c = 2 μ ln ( v u ) (2.113) u v = ( 2 μ r ) e ( r 2 μ ) / 2 μ , t = x 0 c = 2 μ ln v u {:(2.113)uv=(2mu-r)e^((r-2mu)//2mu)","quad t=(x^(0))/(c)=2mu ln(-(v)/(u)):}\begin{equation*} u v=(2 \mu-r) e^{(r-2 \mu) / 2 \mu}, \quad t=\frac{x^{0}}{c}=2 \mu \ln \left(-\frac{v}{u}\right) \tag{2.113} \end{equation*}(2.113)uv=(2μr)e(r2μ)/2μ,t=x0c=2μln(vu)
Here u < 0 u < 0 u < 0u<0u<0 and v > 0 v > 0 v > 0v>0v>0, and we use units with c = 1 c = 1 c=1c=1c=1.
Problem 2.108 Show that a spaceship cannot get out from the black hole region u > 0 u > 0 u > 0u>0u>0 and v > 0 v > 0 v > 0v>0v>0 in Kruskal-Szekeres coordinates.
Problem 2.109 An observer is freely falling to the true singularity r = 0 r = 0 r=0r=0r=0 of a Schwarzschild black hole. We assume that the fall follows a radial ray d Ω = 0 d Ω = 0 d Omega=0d \Omega=0dΩ=0. Since the standard (spherical) coordinates become singular at the Schwarzschild event horizon r = 2 G M / c 2 2 μ r = 2 G M / c 2 2 μ r=2GM//c^(2)-=2mur=2 G M / c^{2} \equiv 2 \mur=2GM/c22μ, we express the initial condition of the observer in terms of the Kruskal-Szekeres coordinates
(2.114) t = x 0 c = 2 μ ln ( v u ) , u v = ( 2 μ r ) e r 2 μ 2 μ , (2.114) t = x 0 c = 2 μ ln v u , u v = ( 2 μ r ) e r 2 μ 2 μ , {:(2.114)t=(x^(0))/(c)=2mu ln((v)/(u))","quad uv=(2mu-r)e^((r-2mu)/(2mu))",":}\begin{equation*} t=\frac{x^{0}}{c}=2 \mu \ln \left(\frac{v}{u}\right), \quad u v=(2 \mu-r) e^{\frac{r-2 \mu}{2 \mu}}, \tag{2.114} \end{equation*}(2.114)t=x0c=2μln(vu),uv=(2μr)er2μ2μ,
for u , v > 0 u , v > 0 u,v > 0u, v>0u,v>0. The initial conditions are
(2.115) u ( 0 ) = 0 , v ( 0 ) = v 0 , u ˙ ( 0 ) = E c v 0 (2.115) u ( 0 ) = 0 , v ( 0 ) = v 0 , u ˙ ( 0 ) = E c v 0 {:(2.115)u(0)=0","quad v(0)=v_(0)","quadu^(˙)(0)=(E)/(cv_(0)):}\begin{equation*} u(0)=0, \quad v(0)=v_{0}, \quad \dot{u}(0)=\frac{E}{c v_{0}} \tag{2.115} \end{equation*}(2.115)u(0)=0,v(0)=v0,u˙(0)=Ecv0
and v ˙ ( 0 ) v ˙ ( 0 ) v^(˙)(0)\dot{v}(0)v˙(0) is determined by the requirement of the 4 -velocity being future-directed with modulus one. Compute the proper time Δ s Δ s Delta s\Delta sΔs for the fall, expressed as an integral
0 2 μ f ( r ) d r 0 2 μ f ( r ) d r int_(0)^(2mu)f(r)dr\int_{0}^{2 \mu} f(r) d r02μf(r)dr
for a certain function f ( r ) f ( r ) f(r)f(r)f(r) of the radius r r rrr expressed in terms of E E EEE and v 0 v 0 v_(0)v_{0}v0.
Hint: From the equations of motion, it can be shown that the quantity
u ˙ v v ˙ u r exp ( 2 μ r 2 μ ) u ˙ v v ˙ u r exp 2 μ r 2 μ ((u^(˙))v-(v^(˙))u)/(r)exp((2mu-r)/(2mu))\frac{\dot{u} v-\dot{v} u}{r} \exp \left(\frac{2 \mu-r}{2 \mu}\right)u˙vv˙urexp(2μr2μ)
is a constant of motion.
Problem 2.110 Consider a Schwarzschild black hole with metric
(2.116) d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 d Ω 2 (2.116) d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d Ω 2 {:(2.116)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.116} \end{equation*}(2.116)ds2=(1rr)dt2(1rr)1dr2r2dΩ2
a) What conclusion can you draw about the singularity r = 0 r = 0 r=0r=0r=0 of this metric from the fact that
(2.117) R μ v α β R μ v α β = 3 r 2 r 6 ? (2.117) R μ v α β R μ v α β = 3 r 2 r 6 ? {:(2.117)R_(mu v alpha beta)R^(mu v alpha beta)=(3r_(**)^(2))/(r^(6))?:}\begin{equation*} R_{\mu v \alpha \beta} R^{\mu v \alpha \beta}=\frac{3 r_{*}^{2}}{r^{6}} ? \tag{2.117} \end{equation*}(2.117)RμvαβRμvαβ=3r2r6?
Do not forget to motivate your answer.
b) Determine what radial light cones look like using the Schwarzschild metric, i.e., how t t ttt depends on r r rrr.
c) Instead of using the Schwarzschild coordinates to describe black holes, it can be useful to use Kruskal-Szekeres coordinates. The metric then takes the form
(2.118) d s 2 = 4 r 3 r e r / r ( d V 2 d U 2 ) r 2 d Ω 2 (2.118) d s 2 = 4 r 3 r e r / r d V 2 d U 2 r 2 d Ω 2 {:(2.118)ds^(2)=(4r_(**)^(3))/(r)*e^(-r//r_(**))(dV^(2)-dU^(2))-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\frac{4 r_{*}^{3}}{r} \cdot e^{-r / r_{*}}\left(d V^{2}-d U^{2}\right)-r^{2} d \Omega^{2} \tag{2.118} \end{equation*}(2.118)ds2=4r3rer/r(dV2dU2)r2dΩ2
Determine what radial light cones look like in these coordinates.

2.8 Weak Field Approximation and Newtonian Limit

Problem 2.111 a) What are the equations of motion for a massive particle in a gravitational potential according to Newton's mechanics and general relativity, respectively? Derive the former from the latter in the Newtonian limit.
b) What are tidal forces in Newton's theory of gravity? How are they related to the gravitational potential? Why are solar tidal forces slightly weaker than lunar tidal forces? In general relativity, explain how the tidal forces are identified with the curvature of spacetime.
Problem 2.112 a) Compute the Ricci tensor R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν in the linear approximation for a metric g μ ν = η μ ν + h μ ν g μ ν = η μ ν + h μ ν g_(mu nu)=eta_(mu nu)+h_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}gμν=ημν+hμν, i.e., you can ignore all but the first-order terms in h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν.
b) Show that a coordinate transformation
(2.119) x μ x μ + χ μ ( x ) , | ν χ μ | 1 , (2.119) x μ x μ + χ μ ( x ) , ν χ μ 1 , {:(2.119)x^(mu)|->x^(mu)+chi^(mu)(x)","quad|del_(nu)chi^(mu)|≪1",":}\begin{equation*} x^{\mu} \mapsto x^{\mu}+\chi^{\mu}(x), \quad\left|\partial_{\nu} \chi^{\mu}\right| \ll 1, \tag{2.119} \end{equation*}(2.119)xμxμ+χμ(x),|νχμ|1,
in the linear approximation described in a) corresponds to a gauge transformation of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν given by
(2.120) h μ ν h μ ν μ χ ν ν χ μ (2.120) h μ ν h μ ν μ χ ν ν χ μ {:(2.120)h_(mu nu)|->h_(mu nu)-del_(mu)chi_(nu)-del_(nu)chi_(mu):}\begin{equation*} h_{\mu \nu} \mapsto h_{\mu \nu}-\partial_{\mu} \chi_{\nu}-\partial_{\nu} \chi_{\mu} \tag{2.120} \end{equation*}(2.120)hμνhμνμχννχμ
c) Show that, by imposing the gauge condition
(2.121) μ h ¯ μ ν = 0 , where h ¯ μ ν = h μ ν h 2 η μ ν , h = η μ ν h μ ν (2.121) μ h ¯ μ ν = 0 ,  where  h ¯ μ ν = h μ ν h 2 η μ ν , h = η μ ν h μ ν {:(2.121)del^(mu) bar(h)_(mu nu)=0","quad" where " bar(h)_(mu nu)=h_(mu nu)-(h)/(2)*eta_(mu nu)","h=eta^(mu nu)h_(mu nu):}\begin{equation*} \partial^{\mu} \bar{h}_{\mu \nu}=0, \quad \text { where } \bar{h}_{\mu \nu}=h_{\mu \nu}-\frac{h}{2} \cdot \eta_{\mu \nu}, h=\eta^{\mu \nu} h_{\mu \nu} \tag{2.121} \end{equation*}(2.121)μh¯μν=0, where h¯μν=hμνh2ημν,h=ημνhμν
the linearized Einstein equations R μ ν = 0 R μ ν = 0 R_(mu nu)=0R_{\mu \nu}=0Rμν=0 reduce to the wave equation for h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν.
Problem 2.113 The spacetime metric corresponding to a weak gravitational potential | Φ ( x ) | c 2 | Φ ( x ) | c 2 |Phi(x)|≪c^(2)|\Phi(\mathbf{x})| \ll c^{2}|Φ(x)|c2 is
(2.122) d s 2 = [ c 2 2 Φ ( x ) ] d t 2 [ 1 + 2 Φ ( x ) c 2 ] ( d x 2 + d y 2 + d z 2 ) (2.122) d s 2 = c 2 2 Φ ( x ) d t 2 1 + 2 Φ ( x ) c 2 d x 2 + d y 2 + d z 2 {:(2.122)ds^(2)=[c^(2)-2Phi(x)]dt^(2)-[1+(2Phi(x))/(c^(2))](dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} d s^{2}=\left[c^{2}-2 \Phi(\mathbf{x})\right] d t^{2}-\left[1+\frac{2 \Phi(\mathbf{x})}{c^{2}}\right]\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{2.122} \end{equation*}(2.122)ds2=[c22Φ(x)]dt2[1+2Φ(x)c2](dx2+dy2+dz2)
a) Find the geodesic equation for this metric in the nonrelativistic and weak field (where you only keep the lowest-order terms in Φ Φ Phi\PhiΦ ) limits. Discuss your result.
b) Compute the redshift of a photon with angular frequency ω ω omega\omegaω moving in the Earth's gravitational field Φ ( x ) = g z Φ ( x ) = g z Phi(x)=-gz\Phi(\mathbf{x})=-g zΦ(x)=gz (independent of x x xxx and y y yyy ) from z = 0 z = 0 z=0z=0z=0 to z = h > 0 z = h > 0 z=h > 0z=h>0z=h>0 as observed by a stationary observer. You should give a detailed derivation of your result. Discuss also how one can understand the result using the equivalence principle.
Problem 2.114 Consider two massive particles moving freely on two close paths on a curved spacetime with metric d s 2 = g μ ν d x μ d x ν d s 2 = g μ ν d x μ d x ν ds^(2)=g_(mu nu)dx^(mu)dx^(nu)d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}ds2=gμνdxμdxν and assume that the positions of these two particles at proper time τ τ tau\tauτ are x μ ( τ ) x μ ( τ ) x^(mu)(tau)x^{\mu}(\tau)xμ(τ) and x μ ( τ ) + s μ ( τ ) x μ ( τ ) + s μ ( τ ) x^(mu)(tau)+s^(mu)(tau)x^{\mu}(\tau)+s^{\mu}(\tau)xμ(τ)+sμ(τ), respectively, with s μ s μ s^(mu)s^{\mu}sμ small (i.e., only terms linear in s μ s μ s^(mu)s^{\mu}sμ need to be taken into account and higherorder terms can be ignored).
a) Derive the geodesic deviation equation
(2.123) D 2 s μ D τ 2 = R α ν β μ s ν d x α d τ d x β d τ (2.123) D 2 s μ D τ 2 = R α ν β μ s ν d x α d τ d x β d τ {:(2.123)(D^(2)s^(mu))/(Dtau^(2))=-R_(alpha nu beta)^(mu)s^(nu)(dx^(alpha))/(d tau)(dx^(beta))/(d tau):}\begin{equation*} \frac{D^{2} s^{\mu}}{D \tau^{2}}=-R_{\alpha \nu \beta}^{\mu} s^{\nu} \frac{d x^{\alpha}}{d \tau} \frac{d x^{\beta}}{d \tau} \tag{2.123} \end{equation*}(2.123)D2sμDτ2=Rανβμsνdxαdτdxβdτ
b) Show that in the Newtonian limit the geodesic deviation equation reduces to the equation for tidal acceleration in Newton's theory of gravitation, i.e.,
(2.124) d 2 s i d t 2 = 2 Φ x i x j s j (2.124) d 2 s i d t 2 = 2 Φ x i x j s j {:(2.124)(d^(2)s^(i))/(dt^(2))=-(del^(2)Phi)/(delx^(i)delx^(j))s^(j):}\begin{equation*} \frac{d^{2} s^{i}}{d t^{2}}=-\frac{\partial^{2} \Phi}{\partial x^{i} \partial x^{j}} s^{j} \tag{2.124} \end{equation*}(2.124)d2sidt2=2Φxixjsj
Hint: Recall that
(2.125) D V μ D τ = d V μ d τ + Γ α β μ d x α d τ V β (2.125) D V μ D τ = d V μ d τ + Γ α β μ d x α d τ V β {:(2.125)(DV^(mu))/(D tau)=(dV^(mu))/(d tau)+Gamma_(alpha beta)^(mu)(dx^(alpha))/(d tau)V^(beta):}\begin{equation*} \frac{D V^{\mu}}{D \tau}=\frac{d V^{\mu}}{d \tau}+\Gamma_{\alpha \beta}^{\mu} \frac{d x^{\alpha}}{d \tau} V^{\beta} \tag{2.125} \end{equation*}(2.125)DVμDτ=dVμdτ+ΓαβμdxαdτVβ
To derive the equation in a) it is convenient, but not necessary, to work in a local inertial frame.
Problem 2.115 Derive a relativistic generalization of the centrifugal force as follows: Consider the motion of a free particle in Minkowski space in a coordinate system rotating with constant angular velocity ω ω omega\omegaω around the z z zzz-coordinate axis.
a) Compute the Christoffel symbols and the geodesic equations in this coordinate system.
Hint: It is convenient to use cylindrical coordinates ( r , φ , z ) ( r , φ , z ) (r,varphi,z)(r, \varphi, z)(r,φ,z).
b) From your result in a), derive the equations of motion in this rotating frame in the nonrelativistic limit (for ω r 1 ω r 1 omega r≪1\omega r \ll 1ωr1 ).
Problem 2.116 a) Find the trajectory of a planet with mass m m mmm moving on a circle in the gravitational potential V ( r ) = G M m / | r | V ( r ) = G M m / | r | V(r)=-GMm//|r|V(\mathbf{r})=-G M m /|\mathbf{r}|V(r)=GMm/|r|, according to Newton's mechanics.
b) There is a natural generalization of the trajectory in a) to general relativity. Explain what this generalization is. Find this generalized trajectory.
Hint: The trajectory can be computed from Hamilton's principle
(2.126) δ ( 1 2 m r ˙ 2 + G M m r ) d t = 0 (2.126) δ 1 2 m r ˙ 2 + G M m r d t = 0 {:(2.126)delta int((1)/(2)mr^(˙)^(2)+(GMm)/(r))dt=0:}\begin{equation*} \delta \int\left(\frac{1}{2} m \dot{\mathbf{r}}^{2}+\frac{G M m}{r}\right) d t=0 \tag{2.126} \end{equation*}(2.126)δ(12mr˙2+GMmr)dt=0
using spherical coordinates ( r , θ , φ ) ( r , θ , φ ) (r,theta,varphi)(r, \theta, \varphi)(r,θ,φ) and assuming θ ( t ) = π / 2 θ ( t ) = π / 2 theta(t)=pi//2\theta(t)=\pi / 2θ(t)=π/2 and r ( t ) = r 0 = r ( t ) = r 0 = r(t)=r_(0)=r(t)=r_{0}=r(t)=r0= constant. Recall that d x 2 + d y 2 + d z 2 = d r 2 + r 2 ( d θ 2 + sin 2 θ d φ 2 ) d x 2 + d y 2 + d z 2 = d r 2 + r 2 d θ 2 + sin 2 θ d φ 2 dx^(2)+dy^(2)+dz^(2)=dr^(2)+r^(2)(dtheta^(2)+sin^(2)theta dvarphi^(2))d x^{2}+d y^{2}+d z^{2}=d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \varphi^{2}\right)dx2+dy2+dz2=dr2+r2(dθ2+sin2θdφ2). Find the relation between r r rrr and the angular momentum L = m r 2 φ ˙ L = m r 2 φ ˙ L=mr^(2)varphi^(˙)L=m r^{2} \dot{\varphi}L=mr2φ˙.
Problem 2.117 Consider the Einstein field equations.
a) What three approximations should be made to obtain the Newtonian limit?
b) Show that, in the Newtonian limit, the Einstein field equations reduce to
(2.127) 2 Φ ρ (2.127) 2 Φ ρ {:(2.127)grad^(2)Phi prop rho:}\begin{equation*} \nabla^{2} \Phi \propto \rho \tag{2.127} \end{equation*}(2.127)2Φρ
where Φ Φ Phi\PhiΦ is the gravitational potential and ρ ρ rho\rhoρ the mass density.
Problem 2.118 Consider a satellite in circular orbit around the Earth at a distance R 1 R 1 R_(1)R_{1}R1 from the surface. The metric outside of the Earth can be considered to be
(2.128) d s 2 = ( 1 + 2 Φ ) d t 2 ( 1 + 2 Φ ) 1 d r 2 r 2 d Ω 2 (2.128) d s 2 = ( 1 + 2 Φ ) d t 2 ( 1 + 2 Φ ) 1 d r 2 r 2 d Ω 2 {:(2.128)ds^(2)=(1+2Phi)dt^(2)-(1+2Phi)^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=(1+2 \Phi) d t^{2}-(1+2 \Phi)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{2.128} \end{equation*}(2.128)ds2=(1+2Φ)dt2(1+2Φ)1dr2r2dΩ2
where Φ = G M / r Φ = G M / r Phi=-GM//r\Phi=-G M / rΦ=GM/r is the classical gravitational potential and d Ω 2 = d θ 2 + d Ω 2 = d θ 2 + dOmega^(2)=dtheta^(2)+d \Omega^{2}=d \theta^{2}+dΩ2=dθ2+ sin 2 θ d ϕ 2 sin 2 θ d ϕ 2 sin^(2)theta dphi^(2)\sin ^{2} \theta d \phi^{2}sin2θdϕ2. What is the eigentime required for the satellite to complete a full orbit around the Earth? How does this compare with the global time t t ttt required for the same orbit?

2.9 Gravitational Lensing

Problem 2.119 Consider a spherical body with radius R 0 R 0 R_(0)R_{0}R0, constant density, and total mass M 0 M 0 M_(0)M_{0}M0. Neutrinos traveling through this body have such small masses that their worldlines can be roughly approximated as null geodesics. Find an expression for the angular deflection of a neutrino with an impact parameter (smallest distance to the body's center) b < R 0 b < R 0 b < R_(0)b<R_{0}b<R0. Verify that your expression has the expected limit when b R 0 b R 0 b rarrR_(0)b \rightarrow R_{0}bR0. You may work in the low-velocity and weak field limits for computing the metric.
Problem 2.120 For large distances from the center of the halo, the Navarro-FrenkWhite (NFW) dark matter halo profile assumes that the matter density varies as ρ ( r ) = k / r 3 ρ ( r ) = k / r 3 rho(r)=k//r^(3)\rho(r)=k / r^{3}ρ(r)=k/r3. Find the deflection angle α α alpha\alphaα due to gravitational lensing of light that passes such a halo at a minimum distance r 0 r 0 r_(0)r_{0}r0. You may assume that the NFW density profile is valid from some radius r = r s < r 0 r = r s < r 0 r=r_(s) < r_(0)r=r_{s}<r_{0}r=rs<r0 and that the mass inside this radius is given by M 0 M 0 M_(0)M_{0}M0.

2.10 Frequency Shifts

Problem 2.121 a) Derive the formula for the gravitational redshift between stationary observers in a static spacetime with line element such that the metric components g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν are independent of a global time coordinate t t ttt.
b) Explain the origin of the gravitational redshift in the case of the Schwarzschild metric and derive the approximative formula
(2.129) z λ B λ A λ A G M c 2 r A (2.129) z λ B λ A λ A G M c 2 r A {:(2.129)z-=(lambda_(B)-lambda_(A))/(lambda_(A))≃(GM)/(c^(2)r_(A)):}\begin{equation*} z \equiv \frac{\lambda_{B}-\lambda_{A}}{\lambda_{A}} \simeq \frac{G M}{c^{2} r_{A}} \tag{2.129} \end{equation*}(2.129)zλBλAλAGMc2rA
for the redshift observed far away at B B BBB, from a source at a radial coordinate r = r A r = r A r=r_(A)r=r_{A}r=rA.
Problem 2.122 A 1+1-dimensional universe is defined as the surface
(2.130) ( c t ) 2 x 2 y 2 = K 2 ( where K > 0 ) (2.130) ( c t ) 2 x 2 y 2 = K 2 (  where  K > 0 ) {:(2.130)(ct)^(2)-x^(2)-y^(2)=-K^(2)quad(" where "K > 0):}\begin{equation*} (c t)^{2}-x^{2}-y^{2}=-K^{2} \quad(\text { where } K>0) \tag{2.130} \end{equation*}(2.130)(ct)2x2y2=K2( where K>0)
in R 3 R 3 R^(3)\mathbb{R}^{3}R3. The metric on the surface is induced by the Minkowski metric d s 2 = c 2 d t 2 d s 2 = c 2 d t 2 ds^(2)=c^(2)dt^(2)-d s^{2}=c^{2} d t^{2}-ds2=c2dt2 d x 2 d y 2 d x 2 d y 2 dx^(2)-dy^(2)d x^{2}-d y^{2}dx2dy2 in R 3 R 3 R^(3)\mathbb{R}^{3}R3. Analyze the frequency shift in this mini-universe between comoving observers.
Problem 2.123 A spaceship is moving radially toward a center of mass M M MMM with a coordinate velocity d r / d t = 0.1 c d r / d t = 0.1 c dr//dt=-0.1 cd r / d t=-0.1 cdr/dt=0.1c, where t t ttt is the Schwarzschild universal time and c 3 10 8 m / s 2 c 3 10 8 m / s 2 c≃3*10^(8)m//s^(2)c \simeq 3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}^{2}c3108 m/s2. An observer in the spaceship is measuring the wavelength of a light signal from a distant star at rest. The light signal travels along the same radius as the observer. The wavelength at r r r rarr oor \rightarrow \inftyr is assumed to be 4000 4000 4000"Å"4000 \AA4000. What is the observed wavelength when G M = 10 20 m 3 / s 2 G M = 10 20 m 3 / s 2 GM=10^(20)m^(3)//s^(2)G M=10^{20} \mathrm{~m}^{3} / \mathrm{s}^{2}GM=1020 m3/s2 and r = 10 6 m r = 10 6 m r=10^(6)mr=10^{6} \mathrm{~m}r=106 m ?
Problem 2.124 Compute the redshift of starlight emitted from the surface of a star with r star = 7 10 8 m r star  = 7 10 8 m r_("star ")=7*10^(8)mr_{\text {star }}=7 \cdot 10^{8} \mathrm{~m}rstar =7108 m and mass M = 2 10 30 kg M = 2 10 30 kg M=2*10^(30)kgM=2 \cdot 10^{30} \mathrm{~kg}M=21030 kg. Use the approximate values G 6.67 10 11 m 3 / ( kg s 2 ) G 6.67 10 11 m 3 / kg s 2 G~~6.67*10^(-11)m^(3)//(kg*s^(2))G \approx 6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right)G6.671011 m3/(kgs2) and c 3 10 8 m / s c 3 10 8 m / s c~~3*10^(8)m//sc \approx 3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}c3108 m/s.
Problem 2.125 Elements in the chromosphere of the Sun emit sharp spectral lines. A student in relativity theory observes one such known spectral line in a spectrometer on Earth. According to general relativity, the emitted light is affected by the mass of the Sun. Calculate, using the general theory of relativity and to lowest order in the gravitational constant, the magnitude and sign of the relative frequency shift Δ v / v Δ v / v Delta v//v\Delta v / vΔv/v of this spectral line. The solar mass is about 2.0 10 30 kg 2.0 10 30 kg 2.0*10^(30)kg2.0 \cdot 10^{30} \mathrm{~kg}2.01030 kg, Newton's gravitational constant is G 6.7 10 11 m 3 / ( kg s 2 ) G 6.7 10 11 m 3 / kg s 2 G~~6.7*10^(-11)m^(3)//(kg*s^(2))G \approx 6.7 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right)G6.71011 m3/(kgs2), the speed of light is c 3.0 10 8 m / s c 3.0 10 8 m / s c~~3.0*10^(8)m//sc \approx 3.0 \cdot 10^{8} \mathrm{~m} / \mathrm{s}c3.0108 m/s, the solar radius is about 7.8 10 8 m 7.8 10 8 m 7.8*10^(8)m7.8 \cdot 10^{8} \mathrm{~m}7.8108 m, and the average distance Sun-Earth is about 1.5 10 11 m 1.5 10 11 m 1.5*10^(11)m1.5 \cdot 10^{11} \mathrm{~m}1.51011 m.
Problem 2.126 A spaceship is launched from the ground station on Earth and is moving radially upward. When it is at an altitude of 1000 km , its velocity is only about 0.1 km / s 0.1 km / s 0.1km//s0.1 \mathrm{~km} / \mathrm{s}0.1 km/s. At that moment, a light signal is sent from the spaceship and is observed at the ground station. Compute the red/blue shifts of the signal from the two most important physical effects. Newton's gravitational constant is G 6.67 G 6.67 G~~6.67G \approx 6.67G6.67. 10 11 m 3 / ( kg s 2 ) 10 11 m 3 / kg s 2 10^(-11)m^(3)//(kg*s^(2))10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right)1011 m3/(kgs2) and the radius and the mass of the Earth are R 6.3 10 3 km R 6.3 10 3 km R~~6.3*10^(3)kmR \approx 6.3 \cdot 10^{3} \mathrm{~km}R6.3103 km and M 5.98 10 24 kg M 5.98 10 24 kg M~~5.98*10^(24)kgM \approx 5.98 \cdot 10^{24} \mathrm{~kg}M5.981024 kg, respectively.
Problem 2.127 Compute the blueshift of a light signal sent from a very distant spaceship and observed at the Earth. Assume that the spaceship is stationary in an approximately static spacetime. Useful information: The distance between the Sun and the Earth is approximately 1.5 10 11 m 1.5 10 11 m 1.5*10^(11)m1.5 \cdot 10^{11} \mathrm{~m}1.51011 m, the speed of light is c 3 10 8 m / s c 3 10 8 m / s c≃3*10^(8)m//sc \simeq 3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}c3108 m/s, G M 1.3 10 20 m 3 / s 2 G M 1.3 10 20 m 3 / s 2 GM_(o.)≃1.3*10^(20)m^(3)//s^(2)G M_{\odot} \simeq 1.3 \cdot 10^{20} \mathrm{~m}^{3} / \mathrm{s}^{2}GM1.31020 m3/s2 and the gravitational potential of the Earth on its surface (normalized to zero at infinity) is 6.24 10 7 m 2 / s 2 6.24 10 7 m 2 / s 2 -6.24*10^(7)m^(2)//s^(2)-6.24 \cdot 10^{7} \mathrm{~m}^{2} / \mathrm{s}^{2}6.24107 m2/s2.
Problem 2.128 A free-falling observer is moving radially away from a black hole with a local velocity that is just large enough to escape the gravitational pull. The free-falling observer is emitting light signals radially toward an observer stationary at infinity. Compute the frequency of the light signal received by the second observer if it was emitted with frequency f 0 f 0 f_(0)f_{0}f0 at radius r r rrr by the first observer.
Problem 2.129 In the two-dimensional spacetime introduced in Problem 2.106, a series of light signals is emitted from a free-falling observer starting at x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0 and received by an observer with a worldline x = x 1 x = x 1 x=x_(1)x=x_{1}x=x1, where x 1 x 1 x_(1)x_{1}x1 is a constant. Find the redshift z z zzz of the light signals as a function of the position x = x e x = x e x=x_(e)x=x_{e}x=xe where the signal was emitted by the free-falling observer.
Problem 2.130 Consider the Robertson-Walker metric written as
(2.131) d s 2 = c 2 d t 2 S ( t ) 2 [ d r 2 1 k r 2 + r 2 d Ω 2 ] (2.131) d s 2 = c 2 d t 2 S ( t ) 2 d r 2 1 k r 2 + r 2 d Ω 2 {:(2.131)ds^(2)=c^(2)dt^(2)-S(t)^(2)[(dr^(2))/(1-kr^(2))+r^(2)dOmega^(2)]:}\begin{equation*} d s^{2}=c^{2} d t^{2}-S(t)^{2}\left[\frac{d r^{2}}{1-k r^{2}}+r^{2} d \Omega^{2}\right] \tag{2.131} \end{equation*}(2.131)ds2=c2dt2S(t)2[dr21kr2+r2dΩ2]
for some fixed parameter k > 0 k > 0 k > 0k>0k>0. We project the metric to two dimensions by setting d Ω = 0 d Ω = 0 d Omega=0d \Omega=0dΩ=0. An observer A A AAA, located at ( t 0 , r 0 t 0 , r 0 t_(0),r_(0)t_{0}, r_{0}t0,r0 ) and at rest with respect to the coordinate r r rrr, sends a light signal. Another observer, located at ( t 1 , r 1 ) t 1 , r 1 (t_(1),r_(1))\left(t_{1}, r_{1}\right)(t1,r1) and also at rest with respect to r r rrr, receives the light signal. After a short time ϵ , A ϵ , A epsilon,A\epsilon, Aϵ,A sends another light signal, which is received by B B BBB at the time t 1 + ϵ t 1 + ϵ t_(1)+epsilon^(')t_{1}+\epsilon^{\prime}t1+ϵ. Compute the ratio ϵ / ϵ ϵ / ϵ epsilon^(')//epsilon\epsilon^{\prime} / \epsilonϵ/ϵ in terms of the unknown function S ( t ) S ( t ) S(t)S(t)S(t) and deduce from this the cosmological redshift.
Problem 2.131 The Robertson-Walker metric (for k = 1 k = 1 k=1k=1k=1 ) can be written as
(2.132) d s 2 = c 2 d t 2 S ( t ) 2 [ d χ 2 + sin 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] (2.132) d s 2 = c 2 d t 2 S ( t ) 2 d χ 2 + sin 2 χ d θ 2 + sin 2 θ d ϕ 2 {:(2.132)ds^(2)=c^(2)dt^(2)-S(t)^(2)[dchi^(2)+sin^(2)chi(dtheta^(2)+sin^(2)theta dphi^(2))]:}\begin{equation*} d s^{2}=c^{2} d t^{2}-S(t)^{2}\left[d \chi^{2}+\sin ^{2} \chi\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right] \tag{2.132} \end{equation*}(2.132)ds2=c2dt2S(t)2[dχ2+sin2χ(dθ2+sin2θdϕ2)]
where 0 χ π / 2 , 0 θ π 0 χ π / 2 , 0 θ π 0 <= chi <= pi//2,0 <= theta <= pi0 \leq \chi \leq \pi / 2,0 \leq \theta \leq \pi0χπ/2,0θπ, and 0 ϕ < 2 π 0 ϕ < 2 π 0 <= phi < 2pi0 \leq \phi<2 \pi0ϕ<2π. Derive the differential equations for the geodesics.

2.11 Gravitational Waves

Problem 2.132 Consider the weak field limit in the harmonic gauge, where g a b = g a b = g_(ab)=g_{a b}=gab= η a b + ε h a b η a b + ε h a b eta_(ab)+epsih_(ab)\eta_{a b}+\varepsilon h_{a b}ηab+εhab and g a b a b χ c = 0 g a b a b χ c = 0 g^(ab)grad_(a)grad_(b)chi^(c)=0g^{a b} \nabla_{a} \nabla_{b} \chi^{c}=0gababχc=0 for the coordinates χ a χ a chi^(a)\chi^{a}χa.
a) Show that the harmonic gauge leaves a residual gauge freedom χ a χ a chi^(a)|->\chi^{a} \mapstoχa χ a = χ a + ε ξ a χ a = χ a + ε ξ a chi^('a)=chi^(a)+epsixi^(a)\chi^{\prime a}=\chi^{a}+\varepsilon \xi^{a}χa=χa+εξa, which also preserves the weak field approximation, as long as g a b a b ξ c = 0 g a b a b ξ c = 0 g^(ab)grad_(a)grad_(b)xi^(c)=0g^{a b} \nabla_{a} \nabla_{b} \xi^{c}=0gababξc=0 and find an expression for h a b h a b h_(ab)^(')h_{a b}^{\prime}hab in the coordinates χ a χ a chi^('a)\chi^{\prime a}χa.
b) Defining h ¯ a b = h a b g a b h / 2 h ¯ a b = h a b g a b h / 2 bar(h)_(ab)=h_(ab)-g_(ab)h//2\bar{h}_{a b}=h_{a b}-g_{a b} h / 2h¯ab=habgabh/2, where h = h a a h = h a a h=h_(a)^(a)h=h_{a}^{a}h=haa, show that
(2.133) a h ¯ a b = 0 (2.133) a h ¯ a b = 0 {:(2.133)del^(a) bar(h)_(ab)=0:}\begin{equation*} \partial^{a} \bar{h}_{a b}=0 \tag{2.133} \end{equation*}(2.133)ah¯ab=0
in the harmonic gauge to leading order in ε ε epsi\varepsilonε.
c) A gravitational plane wave satisfies h ¯ a b = A a b exp ( i k c χ c ) h ¯ a b = A a b exp i k c χ c bar(h)_(ab)=A_(ab)exp(ik_(c)chi^(c))\bar{h}_{a b}=A_{a b} \exp \left(i k_{c} \chi^{c}\right)h¯ab=Aabexp(ikcχc), where the wave equation can be used to conclude that η a b k a k b = 0 η a b k a k b = 0 eta_(ab)k^(a)k^(b)=0\eta_{a b} k^{a} k^{b}=0ηabkakb=0. Using the residual gauge transformation ξ a = i C a exp ( i k c χ c ) ξ a = i C a exp i k c χ c xi^(a)=iC^(a)exp(ik_(c)chi^(c))\xi^{a}=i C^{a} \exp \left(i k_{c} \chi^{c}\right)ξa=iCaexp(ikcχc), find an expression for the amplitude A a b A a b A_(ab)^(')A_{a b}^{\prime}Aab of the gravitational wave in the χ a χ a chi^('a)\chi^{\prime a}χa coordinates.
d) Choosing k 0 = k 3 = 1 k 0 = k 3 = 1 k_(0)=k_(3)=1k_{0}=k_{3}=1k0=k3=1, compute the values of the constants C a C a C^(a)C^{a}Ca for which A 0 a = A 3 a = 0 A 0 a = A 3 a = 0 A_(0a)^(')=A_(3a)^(')=0A_{0 a}^{\prime}=A_{3 a}^{\prime}=0A0a=A3a=0 and A a a = 0 A a a = 0 A_(a)^('a)=0A_{a}^{\prime a}=0Aaa=0.
Hint: Note that the harmonic gauge condition in itself puts some constraints on the form of A a b A a b A_(ab)A_{a b}Aab.
Problem 2.133 Consider gravitational waves described by perturbations h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν of the metric g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν such that g μ ν = η μ ν + h μ ν g μ ν = η μ ν + h μ ν g_(mu nu)=eta_(mu nu)+h_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}gμν=ημν+hμν, where η μ ν η μ ν eta_(mu nu)\eta_{\mu \nu}ημν is the Minkowski metric and | h μ ν | 1 h μ ν 1 |h_(mu nu)|≪1\left|h_{\mu \nu}\right| \ll 1|hμν|1.
a) In the transverse traceless gauge (TT gauge), the conditions h 0 i = 0 h 0 i = 0 h_(0i)=0h_{0 i}=0h0i=0 and η μ ν h μ ν = 0 η μ ν h μ ν = 0 eta^(mu nu)h_(mu nu)=0\eta^{\mu \nu} h_{\mu \nu}=0ημνhμν=0 hold, which means that the Lorenz gauge condition reduces to ν h ν μ = 0 ν h ν μ = 0 del_(nu)h^(nu)_(mu)=0\partial_{\nu} h^{\nu}{ }_{\mu}=0νhνμ=0. Using the TT gauge conditions, show that the number of independent components of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν is two.
b) Consider two particles at rest at ( x , y , z ) = ( 0 , 0 , 0 ) ( x , y , z ) = ( 0 , 0 , 0 ) (x,y,z)=(0,0,0)(x, y, z)=(0,0,0)(x,y,z)=(0,0,0) and ( x , y , z ) = ( L , 0 , 0 ) ( x , y , z ) = ( L , 0 , 0 ) (x,y,z)=(L,0,0)(x, y, z)=(L, 0,0)(x,y,z)=(L,0,0), respectively. A plus-polarized gravitational wave h + h + h_(+)h_{+}h+of frequency f f fff and amplitude h 0 1 h 0 1 h_(0)≪1h_{0} \ll 1h01 passes by, propagating in the z z zzz-direction, such that
( h μ v ( t , x , y , z ) ) = ( 0 0 0 0 0 h + h × 0 0 h × h + 0 0 0 0 0 ) (2.134) = h 0 sin [ 2 π f ( t z c ) ] ( 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 ) . h μ v ( t , x , y , z ) = 0 0 0 0 0 h + h × 0 0 h × h + 0 0 0 0 0 (2.134) = h 0 sin 2 π f t z c 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 . {:[(h_(mu v)(t,x,y,z))=([0,0,0,0],[0,h_(+),h_(xx),0],[0,h_(xx),-h_(+),0],[0,0,0,0])],[(2.134)=h_(0)sin[2pi f(t-(z)/(c))]([0,0,0,0],[0,1,0,0],[0,0,-1,0],[0,0,0,0]).]:}\begin{align*} \left(h_{\mu v}(t, x, y, z)\right) & =\left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & h_{+} & h_{\times} & 0 \\ 0 & h_{\times} & -h_{+} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \\ & =h_{0} \sin \left[2 \pi f\left(t-\frac{z}{c}\right)\right]\left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) . \tag{2.134} \end{align*}(hμv(t,x,y,z))=(00000h+h×00h×h+00000)(2.134)=h0sin[2πf(tzc)](0000010000100000).
Show that the distance d d ddd measured along the x x xxx-axis between the two particles (i.e., the spatial separation along the equal t t ttt hypersurface), as the wave passes, is given by
(2.135) d = [ 1 1 2 h 0 sin ( 2 π f t ) ] L . (2.135) d = 1 1 2 h 0 sin ( 2 π f t ) L . {:(2.135)d=[1-(1)/(2)h_(0)sin(2pi ft)]L.:}\begin{equation*} d=\left[1-\frac{1}{2} h_{0} \sin (2 \pi f t)\right] L . \tag{2.135} \end{equation*}(2.135)d=[112h0sin(2πft)]L.
Problem 2.134 Assuming a source described by the energy-momentum tensor T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν, the linearized Einstein equations are given by
(2.136) h ¯ μ ν = 16 π T μ ν (2.136) h ¯ μ ν = 16 π T μ ν {:(2.136)◻ bar(h)_(mu nu)=16 piT_(mu nu):}\begin{equation*} \square \bar{h}_{\mu \nu}=16 \pi T_{\mu \nu} \tag{2.136} \end{equation*}(2.136)h¯μν=16πTμν
and the solutions in terms of Green's functions can be written as
h ¯ μ ν ( t , x ) = 4 T μ ν ( t | x x | , x ) | x x | d 3 x 4 r T μ ν ( t r , x ) d 3 x h ¯ μ ν ( t , x ) = 4 T μ ν t x x , x x x d 3 x 4 r T μ ν t r , x d 3 x bar(h)_(mu nu)(t,x)=4int(T_(mu nu)(t-|x-x^(')|,x^(')))/(|x-x^(')|)d^(3)x^(')∼(4)/(r)intT_(mu nu)(t-r,x^('))d^(3)x^(')\bar{h}_{\mu \nu}(t, \mathbf{x})=4 \int \frac{T_{\mu \nu}\left(t-\left|\mathbf{x}-\mathbf{x}^{\prime}\right|, \mathbf{x}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} d^{3} x^{\prime} \sim \frac{4}{r} \int T_{\mu \nu}\left(t-r, \mathbf{x}^{\prime}\right) d^{3} x^{\prime}h¯μν(t,x)=4Tμν(t|xx|,x)|xx|d3x4rTμν(tr,x)d3x,
where r | x | r | x | r-=|x|r \equiv|\mathbf{x}|r|x| is far away from the source. Using the conservation law for the energymomentum tensor, i.e., v T μ ν = 0 v T μ ν = 0 grad_(v)T^(mu nu)=0\nabla_{v} T^{\mu \nu}=0vTμν=0, show that the spatial components are
(2.138) h ¯ i j ( t , x ) 2 r d 2 d t 2 ρ ( t r , x ) x i x j d 3 x (2.138) h ¯ i j ( t , x ) 2 r d 2 d t 2 ρ t r , x x i x j d 3 x {:(2.138) bar(h)_(ij)(t","x)∼(2)/(r)*(d^(2))/(dt^(2))int rho(t-r,x^('))x_(i)^(')x_(j)^(')d^(3)x^('):}\begin{equation*} \bar{h}_{i j}(t, \mathbf{x}) \sim \frac{2}{r} \cdot \frac{d^{2}}{d t^{2}} \int \rho\left(t-r, \mathbf{x}^{\prime}\right) x_{i}^{\prime} x_{j}^{\prime} d^{3} x^{\prime} \tag{2.138} \end{equation*}(2.138)h¯ij(t,x)2rd2dt2ρ(tr,x)xixjd3x
where ρ = T 00 ρ = T 00 rho=T^(00)\rho=T^{00}ρ=T00 is the mass-energy density of the source.
Problem 2.135 a) For a neutron-star binary (with total mass M 2.8 M M 2.8 M M≃2.8M_(o.)M \simeq 2.8 M_{\odot}M2.8M ) at a distance of 5 kpc with orbital period P = 1 h P = 1 h P=1hP=1 \mathrm{~h}P=1 h, estimate the amplitude h h hhh of the gravitational waves.
b) Again, for the same system, but now with P = 0.02 s P = 0.02 s P=0.02sP=0.02 \mathrm{~s}P=0.02 s (giving f GW = 2 f orbit = f GW = 2 f orbit  = f_(GW)=2f_("orbit ")=f_{\mathrm{GW}}=2 f_{\text {orbit }}=fGW=2forbit = 100 Hz , which lies in the sensitive band of the gravitational wave observatory LIGO), estimate h h hhh at a distance of 15 Mpc , which is approximately the distance to the Virgo cluster of galaxies.
c) Estimate the orbital separation R R RRR when P = 0.02 s P = 0.02 s P=0.02sP=0.02 \mathrm{~s}P=0.02 s.
d) For a neutron star at 1 kpc with a nonspherical deformation of mass δ M = δ M = delta M=\delta M=δM= 10 6 M 10 6 M 10^(-6)M_(o.)10^{-6} M_{\odot}106M, a spin frequency of 50 Hz , and a stellar radius of 10 km , determine the gravitational wave amplitude h h hhh at Earth.
Problem 2.136 The first direct observation of gravitational waves was performed on September 14, 2015. This was presented by the LIGO and Virgo Collaborations on February 11, 2016, and was awarded the Nobel Prize in Physics in 2017. The original signal was named GW150914 and it was also the first observation of a binary black hole merger.
a) GW150914 had a maximal amplitude of h 10 21 h 10 21 h≃10^(-21)h \simeq 10^{-21}h1021 at a frequency of f f f≃f \simeqf 200 Hz . Compute the corresponding energy flux at the Earth. The binary source of GW150914 is situated at an estimated distance of about 400 Mpc .
b) Estimate the energy flux in electromagnetic waves that is received at Earth from a full moon. Compare this energy flux to the gravitational wave energy flux of GW150914.
Note:
(2.139) c 5 G 3.63 10 52 W (2.139) c 5 G 3.63 10 52 W {:(2.139)(c^(5))/(G)≃3.63*10^(52)W:}\begin{equation*} \frac{c^{5}}{G} \simeq 3.63 \cdot 10^{52} \mathrm{~W} \tag{2.139} \end{equation*}(2.139)c5G3.631052 W
is equal to 1 in geometric units, i.e., c = 1 c = 1 c=1c=1c=1 and G = 1 G = 1 G=1G=1G=1.

2.12 Cosmology and Friedmann-Lemaître-Robertson-Walker Metric

Problem 2.137 Consider the linearly expanding spacetime with metric d s 2 = d s 2 = ds^(2)=d s^{2}=ds2= d t 2 H 2 t 2 d x 2 d t 2 H 2 t 2 d x 2 dt^(2)-H^(2)t^(2)dx^(2)d t^{2}-H^{2} t^{2} d x^{2}dt2H2t2dx2. You start at t = t 0 , x = 0 t = t 0 , x = 0 t=t_(0),x=0t=t_{0}, x=0t=t0,x=0 and want to arrive at x = L x = L x=Lx=Lx=L at t = t 1 t = t 1 t=t_(1)t=t_{1}t=t1 without accelerating at any time. Find an expression for your position x ( t ) x ( t ) x(t)x(t)x(t) as a function of the global time t t ttt. What is the largest L L LLL you can arrive at in finite global time?
Problem 2.138 In a 2-dimensional mini-universe the metric element is given by
(2.140) d s 2 = c 2 d t 2 S ( t ) 2 d χ 2 (2.140) d s 2 = c 2 d t 2 S ( t ) 2 d χ 2 {:(2.140)ds^(2)=c^(2)dt^(2)-S(t)^(2)dchi^(2):}\begin{equation*} d s^{2}=c^{2} d t^{2}-S(t)^{2} d \chi^{2} \tag{2.140} \end{equation*}(2.140)ds2=c2dt2S(t)2dχ2
where S ( t ) S ( t ) S(t)S(t)S(t) is some positive function of the time t t ttt, constant in the "space" variable χ χ chi\chiχ. Explain the cosmological redshift and derive an expression for it by studying the emission and detection of light signals by comoving observers at two different locations χ 0 χ 0 chi_(0)\chi_{0}χ0 and χ 1 χ 1 chi_(1)\chi_{1}χ1.
Problem 2.139 Consider the two-dimensional de Sitter universe with the metric ( c = 1 c = 1 c=1c=1c=1 )
(2.141) d s 2 = d t 2 e 2 t / R d x 2 (2.141) d s 2 = d t 2 e 2 t / R d x 2 {:(2.141)ds^(2)=dt^(2)-e^(2t//R)dx^(2):}\begin{equation*} d s^{2}=d t^{2}-e^{2 t / R} d x^{2} \tag{2.141} \end{equation*}(2.141)ds2=dt2e2t/Rdx2
where R > 0 R > 0 R > 0R>0R>0 is a constant. Find an expression for the cosmological redshift between comoving observers at x 0 > 0 x 0 > 0 x_(0) > 0x_{0}>0x0>0 and x 1 > x 0 x 1 > x 0 x_(1) > x_(0)x_{1}>x_{0}x1>x0.
Problem 2.140 Consider the two-dimensional de Sitter universe as defined in Problem 2.139.
a) Compute all nonzero Christoffel symbols for this metric.
b) Find the explicit form of the wave equation g μ ν μ ν Φ = 0 g μ ν μ ν Φ = 0 g_(mu nu)grad^(mu)grad^(nu)Phi=0g_{\mu \nu} \nabla^{\mu} \nabla^{\nu} \Phi=0gμνμνΦ=0, where Φ Φ Phi\PhiΦ is a scalar field, in this universe.
Problem 2.141 Consider the Robertson-Walker spacetime described by the metric
(2.142) d s 2 = d t 2 e 2 t / t H [ d r 2 + r 2 ( d θ 2 + sin 2 θ d φ 2 ) ] , (2.142) d s 2 = d t 2 e 2 t / t H d r 2 + r 2 d θ 2 + sin 2 θ d φ 2 , {:(2.142)ds^(2)=dt^(2)-e^(2t//t_(H))[dr^(2)+r^(2)(dtheta^(2)+sin^(2)theta dvarphi^(2))]",":}\begin{equation*} d s^{2}=d t^{2}-e^{2 t / t_{H}}\left[d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \varphi^{2}\right)\right], \tag{2.142} \end{equation*}(2.142)ds2=dt2e2t/tH[dr2+r2(dθ2+sin2θdφ2)],
with the coordinates ( x μ ) = ( t , r , θ , φ ) x μ = ( t , r , θ , φ ) (x^(mu))=(t,r,theta,varphi)\left(x^{\mu}\right)=(t, r, \theta, \varphi)(xμ)=(t,r,θ,φ), where t t ttt is the universal time and t H 14 Gyr t H 14 Gyr t_(H)~~14Gyrt_{H} \approx 14 \mathrm{Gyr}tH14Gyr is the Hubble time.
a) Compute the path r ( t ) r ( t ) r(t)r(t)r(t) of a light pulse emitted at time t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0 at the origin r = 0 r = 0 r=0r=0r=0. You may assume that the light ray moves along the line θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and φ = 0 φ = 0 varphi=0\varphi=0φ=0 so that d r / d t > 0 d r / d t > 0 dr//dt > 0d r / d t>0dr/dt>0.
b) Compute the proper distance between the origin r = 0 r = 0 r=0r=0r=0 and a point r > 0 r > 0 r > 0r>0r>0 on the line θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and φ = 0 φ = 0 varphi=0\varphi=0φ=0 at fixed universal time t t ttt.
c) Compute the spectral shift of the light pulse in a) during the time between emission at the origin and arrival at a point r > 0 r > 0 r > 0r>0r>0.
Hint: The spectral shift is defined as z = λ rec / λ em 1 z = λ rec  / λ em 1 z=lambda_("rec ")//lambda_(em)-1z=\lambda_{\text {rec }} / \lambda_{\mathrm{em}}-1z=λrec /λem1, where λ em λ em lambda_(em)\lambda_{\mathrm{em}}λem is the wavelength of the light at emission and λ rec λ rec  lambda_("rec ")\lambda_{\text {rec }}λrec  its wavelength when it arrives.
Problem 2.142 a) Derive Hubble's law
(2.143) v p = a ˙ ( t ) a ( t ) d p (2.143) v p = a ˙ ( t ) a ( t ) d p {:(2.143)v_(p)=((a^(˙))(t))/(a(t))d_(p):}\begin{equation*} v_{p}=\frac{\dot{a}(t)}{a(t)} d_{p} \tag{2.143} \end{equation*}(2.143)vp=a˙(t)a(t)dp
where a ˙ ( t ) = d a / d t a ˙ ( t ) = d a / d t a^(˙)(t)=da//dt\dot{a}(t)=d a / d ta˙(t)=da/dt and v p v p v_(p)v_{p}vp and d p d p d_(p)d_{p}dp are the proper velocity and the proper distance, respectively, from the Robertson-Walker metric
(2.144) d s 2 = d t 2 a ( t ) 2 G i j d x i d x j (2.144) d s 2 = d t 2 a ( t ) 2 G i j d x i d x j {:(2.144)ds^(2)=dt^(2)-a(t)^(2)G_(ij)dx^(i)dx^(j):}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2} G_{i j} d x^{i} d x^{j} \tag{2.144} \end{equation*}(2.144)ds2=dt2a(t)2Gijdxidxj
b) What are the physical consequences of Hubble's law?
Problem 2.143 a) The first Friedmann equation can be written as
(2.145) a ˙ 2 a 2 + k a 2 = 8 π G 3 ρ , (2.145) a ˙ 2 a 2 + k a 2 = 8 π G 3 ρ , {:(2.145)(a^(˙)^(2))/(a^(2))+(k)/(a^(2))=(8pi G)/(3)rho",":}\begin{equation*} \frac{\dot{a}^{2}}{a^{2}}+\frac{k}{a^{2}}=\frac{8 \pi G}{3} \rho, \tag{2.145} \end{equation*}(2.145)a˙2a2+ka2=8πG3ρ,
which can also be written as 1 Ω = k / a ˙ 2 1 Ω = k / a ˙ 2 1-Omega=-k//a^(˙)^(2)1-\Omega=-k / \dot{a}^{2}1Ω=k/a˙2, where Ω = ρ / ρ c Ω = ρ / ρ c Omega=rho//rho_(c)\Omega=\rho / \rho_{c}Ω=ρ/ρc. Assume now that k k kkk is small compared to the energy density ρ ρ rho\rhoρ, which mainly consists of the cosmological constant ρ Λ ρ Λ rho_(Lambda)\rho_{\Lambda}ρΛ, thus leading to an inflationary universe. Show that the longer this scenario is assumed to last, the closer Ω Ω Omega\OmegaΩ gets to one.
b) Describe the flatness problem in words and how it is solved by inflation.
Problem 2.144 Our present universe can roughly be described by a spatially flat Friedmann-Lemaître-Robertson-Walker spacetime with Ω Λ = 0.7 , Ω m = 0.3 Ω Λ = 0.7 , Ω m = 0.3 Omega_(Lambda)=0.7,Omega_(m)=0.3\Omega_{\Lambda}=0.7, \Omega_{m}=0.3ΩΛ=0.7,Ωm=0.3, Ω r 10 4 Ω r 10 4 Omega_(r)≃10^(-4)\Omega_{r} \simeq 10^{-4}Ωr104, being the density parameters of the cosmological constant (dark energy), matter, and radiation, respectively.
a) How much smaller was the scale factor when the energy density of the dark energy was equal to the energy density of matter?
b) At what redshift z z zzz did the matter-radiation equality (equal amounts of radiation and matter energy density) occur?
Problem 2.145 Our current universe is roughly described by a dark energy component Ω Λ = 0.7 Ω Λ = 0.7 Omega_(Lambda)=0.7\Omega_{\Lambda}=0.7ΩΛ=0.7 and a matter component Ω m = 0.3 Ω m = 0.3 Omega_(m)=0.3\Omega_{m}=0.3Ωm=0.3. Determine an integral expression for the future behavior of the scale factor a ( t ) a ( t ) a(t)a(t)a(t). You may assume that the scale factor today is a 0 = 1 a 0 = 1 a_(0)=1a_{0}=1a0=1 and that the current Hubble parameter is H 0 H 0 H_(0)H_{0}H0. Plot the result of your integral for 0 a ( t ) 100 0 a ( t ) 100 0 <= a(t) <= 1000 \leq a(t) \leq 1000a(t)100. Compare your result to the analytic result a ( t ) = a ( t ) = a(t)=a(t)=a(t)= exp ( H 0 ( t t 0 ) ) exp H 0 t t 0 exp(H_(0)(t-t_(0)))\exp \left(H_{0}\left(t-t_{0}\right)\right)exp(H0(tt0)) for Ω Λ = 1 Ω Λ = 1 Omega_(Lambda)=1\Omega_{\Lambda}=1ΩΛ=1 and Ω m = 0 Ω m = 0 Omega_(m)=0\Omega_{m}=0Ωm=0 and determine the age of the universe t 0 t 0 t_(0)t_{0}t0 if a ( 0 ) = 0 a ( 0 ) = 0 a(0)=0a(0)=0a(0)=0.
Hint: You may use numerical integration.
Problem 2.146 A toy model 1+1-dimensional circular Robertson-Walker universe has the line element
(2.146) d s 2 = d t 2 a ( t ) 2 d φ 2 (2.146) d s 2 = d t 2 a ( t ) 2 d φ 2 {:(2.146)ds^(2)=dt^(2)-a(t)^(2)dvarphi^(2):}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2} d \varphi^{2} \tag{2.146} \end{equation*}(2.146)ds2=dt2a(t)2dφ2
where φ φ varphi\varphiφ and φ + 2 n π ( n N ) φ + 2 n π ( n N ) varphi+2n pi(n inN)\varphi+2 n \pi(n \in \mathbb{N})φ+2nπ(nN) correspond to the same spatial point. An object is thrown from φ = φ 0 φ = φ 0 varphi=varphi_(0)\varphi=\varphi_{0}φ=φ0 at time t 0 t 0 t_(0)t_{0}t0 with a velocity v v vvv relative to a comoving observer.
a) Find a condition that must be satisfied in order for the object to complete a full lap around the universe to reach the comoving observer again from the other direction.
b) What is the relative velocity between the object and a comoving observer at an arbitrary time t t ttt ?
You may assume that the scale factor a ( t ) a ( t ) a(t)a(t)a(t) has a known dependence on t t ttt.
Problem 2.147 Consider a flat universe containing only matter and radiation components such that the radiation density today corresponds to Ω rad = x 1 Ω rad = x 1 Omega_(rad)=x≪1\Omega_{\mathrm{rad}}=x \ll 1Ωrad=x1. Find an expression for the time that has passed since matter and radiation had equal energy densities.
Problem 2.148 Consider a Friedmann-Lemaittre-Robertson-Walker universe with curvature parameter κ 0 κ 0 kappa!=0\kappa \neq 0κ0. Determine the condition on the equation-of-state parameter w = p / ρ w = p / ρ w=p//rhow=p / \rhow=p/ρ such that the curvature parameter | Ω K | = | 1 / ( H a ) 2 | Ω K = 1 / ( H a ) 2 |Omega_(K)|=|-1//(Ha)^(2)|\left|\Omega_{K}\right|=\left|-1 /(H a)^{2}\right||ΩK|=|1/(Ha)2| decreases with cosmological time t t ttt for an expanding universe ( a ˙ > 0 a ˙ > 0 a^(˙) > 0\dot{a}>0a˙>0 ).
Problem 2.149 For a flat single-component Friedmann-Lemaître-Robertson-
Walker universe with an arbitrary, but fixed, equation-of-state parameter w = p / ρ w = p / ρ w=p//rhow=p / \rhow=p/ρ :
a) Compute the scale factor a ( t ) a ( t ) a(t)a(t)a(t) as a function of cosmological time t t ttt.
b) Compute the Hubble parameter H ( t ) H ( t ) H(t)H(t)H(t) as a function of t t ttt.
Hint: You may assume that, for some cosmological time t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0, we normalize our parameters such that a 0 = a ( t 0 ) = 1 a 0 = a t 0 = 1 a_(0)=a(t_(0))=1a_{0}=a\left(t_{0}\right)=1a0=a(t0)=1 and H 0 = H ( t 0 ) H 0 = H t 0 H_(0)=H(t_(0))H_{0}=H\left(t_{0}\right)H0=H(t0) are known. Your answers should be given in terms of ω , t 0 ω , t 0 omega,t_(0)\omega, t_{0}ω,t0, and H 0 H 0 H_(0)H_{0}H0.
Problem 2.150 A scalar field ϕ ϕ phi\phiϕ with potential energy density V ( ϕ ) V ( ϕ ) V(phi)V(\phi)V(ϕ) has a Lagrangian density given by
(2.147) L = 1 2 g μ ν ( μ ϕ ) ( ν ϕ ) V ( ϕ ) . (2.147) L = 1 2 g μ ν μ ϕ ν ϕ V ( ϕ ) . {:(2.147)L=(1)/(2)g^(mu nu)(del_(mu)phi)(del_(nu)phi)-V(phi).:}\begin{equation*} \mathscr{L}=\frac{1}{2} g^{\mu \nu}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \phi\right)-V(\phi) . \tag{2.147} \end{equation*}(2.147)L=12gμν(μϕ)(νϕ)V(ϕ).
a) Derive the equation of motion for the scalar field ϕ ϕ phi\phiϕ.
b) Assuming that the N + 1 N + 1 N+1N+1N+1-dimensional spacetime has a metric given by
(2.148) d s 2 = g μ ν d x μ d x ν = d t 2 a ( t ) 2 G i j d x i d x j (2.148) d s 2 = g μ ν d x μ d x ν = d t 2 a ( t ) 2 G i j d x i d x j {:(2.148)ds^(2)=g_(mu nu)dx^(mu)dx^(nu)=dt^(2)-a(t)^(2)G_(ij)dx^(i)dx^(j):}\begin{equation*} d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}=d t^{2}-a(t)^{2} G_{i j} d x^{i} d x^{j} \tag{2.148} \end{equation*}(2.148)ds2=gμνdxμdxν=dt2a(t)2Gijdxidxj
where G i j G i j G_(ij)G_{i j}Gij are the metric components on an N N NNN-dimensional Riemannian manifold, and that the scalar field ϕ ϕ phi\phiϕ only depends on the time coordinate t t ttt, show that the scalar field is an ideal fluid and find the (time-dependent) equation-of-state parameter w = p / ρ 0 w = p / ρ 0 w=p//rho_(0)w=p / \rho_{0}w=p/ρ0.

  1. 1 1 ^(1){ }^{1}1 Note the abuse of notation - the symbol x 2 x 2 x^(2)x^{2}x2 denotes both the 'length' of the vector x x xxx and the second spatial contravariant component of the vector x x xxx. Unfortunately, this type of abuse of notation is difficult to avoid in relativity theory, since the notation would otherwise be too cumbersome.